TimSort Java源码个人解读
2024-10-21 10:21:14
/*JDK 1.8
*/ package java.util; /**
* A stable, adaptive, iterative mergesort that requires far fewer than
* n lg(n) comparisons when running on partially sorted arrays, while
* offering performance comparable to a traditional mergesort when run
* on random arrays. Like all proper mergesorts, this sort is stable and
* runs O(n log n) time (worst case). In the worst case, this sort requires
* temporary storage space for n/2 object references; in the best case,
* it requires only a small constant amount of space.
*
* This implementation was adapted from Tim Peters's list sort for
* Python, which is described in detail here:
*
* http://svn.python.org/projects/python/trunk/Objects/listsort.txt
*
* Tim's C code may be found here:
*
* http://svn.python.org/projects/python/trunk/Objects/listobject.c
*
* The underlying techniques are described in this paper (and may have
* even earlier origins):
*
* "Optimistic Sorting and Information Theoretic Complexity"
* Peter McIlroy
* SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
* pp 467-474, Austin, Texas, 25-27 January 1993.
*
* While the API to this class consists solely of static methods, it is
* (privately) instantiable; a TimSort instance holds the state of an ongoing
* sort, assuming the input array is large enough to warrant the full-blown
* TimSort. Small arrays are sorted in place, using a binary insertion sort.
*
* @author Josh Bloch
*/
class TimSort<T> {
/**
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
*
* This constant should be a power of two. It was 64 in Tim Peter's C
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* to be a number that's not a power of two, you'll need to change the
* {@link #minRunLength} computation.
*
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
*/
private static final int MIN_MERGE = 32; /**
* The array being sorted.
*/
private final T[] a; /**
* The comparator for this sort.
*/
private final Comparator<? super T> c; /**
* When we get into galloping mode, we stay there until both runs win less
* often than MIN_GALLOP consecutive times.
*/
private static final int MIN_GALLOP = 7; /**
* This controls when we get *into* galloping mode. It is initialized
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
* random data, and lower for highly structured data.
*/
private int minGallop = MIN_GALLOP; /**
* Maximum initial size of tmp array, which is used for merging. The array
* can grow to accommodate demand.
*
* Unlike Tim's original C version, we do not allocate this much storage
* when sorting smaller arrays. This change was required for performance.
*/
private static final int INITIAL_TMP_STORAGE_LENGTH = 256; /**
* Temp storage for merges. A workspace array may optionally be
* provided in constructor, and if so will be used as long as it
* is big enough.
*/
private T[] tmp;
private int tmpBase; // base of tmp array slice
private int tmpLen; // length of tmp array slice /**
* A stack of pending runs yet to be merged. Run i starts at
* address base[i] and extends for len[i] elements. It's always
* true (so long as the indices are in bounds) that:
*
* runBase[i] + runLen[i] == runBase[i + 1]
*
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
*/
private int stackSize = 0; // Number of pending runs on stack
private final int[] runBase;
private final int[] runLen; /**
* Creates a TimSort instance to maintain the state of an ongoing sort.
*
* @param a the array to be sorted
* @param c the comparator to determine the order of the sort
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
*/
private TimSort(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen) {
this.a = a;
this.c = c; // Allocate temp storage (which may be increased later if necessary)
int len = a.length;
// 确定临时数组的长度， 如果低于默认值256的2倍， 则空间大小为原始数组a的长度乘以2， 否则为默认长度
int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
if (work == null || workLen < tlen || workBase + tlen > work.length) {
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
T[] newArray = (T[])java.lang.reflect.Array.newInstance
(a.getClass().getComponentType(), tlen);
tmp = newArray;
tmpBase = 0;
tmpLen = tlen;
}
else {
// 当指定的work数组不为空， 且workLen大于计算出的tlen的长度， 并且work数组的有效长度大于tlen的长度时， 使用指定的临时数组
tmp = work;
tmpBase = workBase;
tmpLen = workLen;
} /*
* Allocate runs-to-be-merged stack (which cannot be expanded). The
* stack length requirements are described in listsort.txt. The C
* version always uses the same stack length (85), but this was
* measured to be too expensive when sorting "mid-sized" arrays (e.g.,
* 100 elements) in Java. Therefore, we use smaller (but sufficiently
* large) stack lengths for smaller arrays. The "magic numbers" in the
* computation below must be changed if MIN_MERGE is decreased. See
* the MIN_MERGE declaration above for more information.
* The maximum value of 49 allows for an array up to length
* Integer.MAX_VALUE-4, if array is filled by the worst case stack size
* increasing scenario. More explanations are given in section 4 of:
* http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf
*/
int stackLen = (len < 120 ? 5 :
len < 1542 ? 10 :
len < 119151 ? 24 : 49);
runBase = new int[stackLen];
runLen = new int[stackLen];
} /*
* The next method (package private and static) constitutes the
* entire API of this class.
*/ /**
* Sorts the given range, using the given workspace array slice
* for temp storage when possible. This method is designed to be
* invoked from public methods (in class Arrays) after performing
* any necessary array bounds checks and expanding parameters into
* the required forms.
*
* @param a the array to be sorted
* @param lo the index of the first element, inclusive, to be sorted
* @param hi the index of the last element, exclusive, to be sorted
* @param c the comparator to use
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
* @since 1.8
*/
static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c,
T[] work, int workBase, int workLen) {
assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; int nRemaining = hi - lo; //总共的待排序的元素个数
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted // If array is small, do a "mini-TimSort" with no merges
// 当元素个数小于7时， 使用折半插入排序， 因为插入排序对于元素个数少的数组更快。
if (nRemaining < MIN_MERGE) {
// initRunLen是初始的有序的元素的个数，而从索引位置lo + initRunLen开始， 后面为乱序的元素。 一种优化方法， lo + initRunLen之前的不用排序了
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
// 折半插入排序
binarySort(a, lo, hi, lo + initRunLen, c);
return;
} /**
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs
* to maintain stack invariant.
*/
// 新建一个TimSort实例， 存储运行时的状态， 比如临时的run(一个run即一个有序的数组)
TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen);
// 查找一个minRun值， 小于minRun时使用折半插入排序，MIN_MERGE/2 <= minRun <= MIN_MERGE
int minRun = minRunLength(nRemaining);
do {
// Identify next run
// 找到下一个有序的数组的长度
int runLen = countRunAndMakeAscending(a, lo, hi, c); // If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
// 使用折半插入排序扩展数组， 使之达到minRun， 因为元素个数小于minRun， 折半插入排序更快速
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen, c);
runLen = force;
} // Push run onto pending-run stack, and maybe merge
// 把这个有序数组压入栈中
ts.pushRun(lo, runLen);
/**
* 判断当
* runLen[i - 3] <= runLen[i - 2] + runLen[i - 1]
* 且 runLen[i-3] < runLen[i-1]时
* 或
* runLen[i - 2] <= runLen[i - 1]
* 合并较小的两个有序数组， 以达到最大的平衡（即每个数组大小基本相同）
*/
ts.mergeCollapse(); // Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0); // Merge all remaining runs to complete sort
assert lo == hi;
//合并剩余数组
ts.mergeForceCollapse();
assert ts.stackSize == 1;
} /**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
*
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* elements from index {@code lo}, inclusive, to {@code start},
* exclusive are already sorted.
*
* @param a the array in which a range is to be sorted
* @param lo the index of the first element in the range to be sorted
* @param hi the index after the last element in the range to be sorted
* @param start the index of the first element in the range that is
* not already known to be sorted ({@code lo <= start <= hi})
* @param c comparator to used for the sort
*/
@SuppressWarnings("fallthrough")
private static <T> void binarySort(T[] a, int lo, int hi, int start,
Comparator<? super T> c) {
assert lo <= start && start <= hi;
// start之前的有序元素直接略过
if (start == lo)
start++;
// 从start到hi， 使用折半插入排序进行数组排序
for ( ; start < hi; start++) {
//待插入的元素
T pivot = a[start]; // Set left (and right) to the index where a[start] (pivot) belongs
// 从left到right， 找到插入位置
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (c.compare(pivot, a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
// left即为最终的插入位置， 因为start>=lo && start <=hi， 所以最终一定会找到一个位置使得pivot>=a[mid]， 因此最终一定是pivot >= right， 因此最终为left的位置， 即mid+1
assert left == right; /*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
switch (n) {
case 2: a[left + 2] = a[left + 1]; // 如果待移动元素个数小于等于2则直接移动
case 1: a[left + 1] = a[left];
break;
default: System.arraycopy(a, left, a, left + 1, n); // 从left开始往后移， 然后把start位置的元素插入到原来的left的位置
}
a[left] = pivot;
}
} /**
* Returns the length of the run beginning at the specified position in
* the specified array and reverses the run if it is descending (ensuring
* that the run will always be ascending when the method returns).
*
* A run is the longest ascending sequence with:
*
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
* or the longest descending sequence with:
*
* a[lo] > a[lo + 1] > a[lo + 2] > ...
*
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely
* reverse a descending sequence without violating stability.
*
* @param a the array in which a run is to be counted and possibly reversed
* @param lo index of the first element in the run
* @param hi index after the last element that may be contained in the run.
It is required that {@code lo < hi}.
* @param c the comparator to used for the sort
* @return the length of the run beginning at the specified position in
* the specified array
*/
private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
Comparator<? super T> c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1; // lo < hi， 且lo + 1 = hi， 因此是有序且升序的， 直接返回 // Find end of run, and reverse range if descending
if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
// 如果是降序的， 找到最长的有序降序序列的长度， 并且把序列倒置， 使之升序
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
// 如果是升序的， 同样找到最长的有序序列的长度
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
} // 返回有序序列的长度
return runHi - lo;
} /**
* Reverse the specified range of the specified array.
*
* @param a the array in which a range is to be reversed
* @param lo the index of the first element in the range to be reversed
* @param hi the index after the last element in the range to be reversed
*/
private static void reverseRange(Object[] a, int lo, int hi) {
hi--;
// 首尾倒置
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
} /**
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2.
* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
* is close to, but strictly less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
// n&1是判断n是能否被2整除， 如果不能被2整除， 最后一个Bit位一定是1,则1&1为1， r = r | 1 为1
r |= (n & 1);
n >>= 1;
}
return n + r;
} /**
* Pushes the specified run onto the pending-run stack.
*
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
*/
// 把有序序列起始位置和长度放入栈中
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
} /**
* Examines the stack of runs waiting to be merged and merges adjacent runs
* until the stack invariants are reestablished:
*
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 2. runLen[i - 2] > runLen[i - 1]
*
* This method is called each time a new run is pushed onto the stack,
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
*/
// 判断合并栈顶的三个元素中较小的两个， 或如果第二个元素比第一个小， 则合并， 使栈中所有的序列大小达到近似相等
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
if (runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
} else if (runLen[n] <= runLen[n + 1]) {
mergeAt(n);
} else {
break; // Invariant is established
}
}
} /**
* Merges all runs on the stack until only one remains. This method is
* called once, to complete the sort.
*/
// 最后合并栈中所有的序列， 直到最后只剩一个有序序列
private void mergeForceCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
// 如果runLen[n-1] < runLen[n+1]， 则合并较小的较小的runBase[n-1]和runBase[n]， 否则合并runBase[n]和runBase[n+1]
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
}
} /**
* Merges the two runs at stack indices i and i+1. Run i must be
* the penultimate or antepenultimate run on the stack. In other words,
* i must be equal to stackSize-2 or stackSize-3.
*
* @param i stack index of the first of the two runs to merge
*/
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
assert i == stackSize - 2 || i == stackSize - 3; int base1 = runBase[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
assert len1 > 0 && len2 > 0;
assert base1 + len1 == base2; /*
* Record the length of the combined runs; if i is the 3rd-last
* run now, also slide over the last run (which isn't involved
* in this merge). The current run (i+1) goes away in any case.
*/
runLen[i] = len1 + len2;
// 如果i是栈顶倒数第三个元素， 则最后i+1一定会合进i数组， 因此i+1的位置替换成i+2
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--; /*
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
*/
// 找到run2的首元素在run1中的位置
int k = gallopRight(a[base2], a, base1, len1, 0, c);
assert k >= 0;
// 忽略k之前的序列， 因为已经有序， 减少比较次数
base1 += k;
len1 -= k;
if (len1 == 0)
return; /*
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
*/
// 找打run1的尾元素在run2中的位置
len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
assert len2 >= 0;
// len2 == 0， 说明run1和run2已经是一个整体有序的序列了， 直接返回。
if (len2 == 0)
return; // Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
else
mergeHi(base1, len1, base2, len2);
} /**
* Locates the position at which to insert the specified key into the
* specified sorted range; if the range contains an element equal to key,
* returns the index of the leftmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
* pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
* In other words, key belongs at index b + k; or in other words,
* the first k elements of a should precede key, and the last n - k
* should follow it.
*/
private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
if (c.compare(key, a[base + hint]) > 0) {
// 查找区间[base + hint, len], 查找使得a[base+hint+lastOfs] < key <= a[base+hint+ofs]成立的ofs值。
int maxOfs = len - hint;
// 向右查找， 最大可能Ofs为maxOfs-1， 即len - hint - 1， 即a[base + len - 1]
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
// 记录上一次ofs的值， 存到lastOfs中
lastOfs = ofs;
// ofs乘以2再加1
ofs = (ofs << 1) + 1;
// 整数溢出
if (ofs <= 0)
ofs = maxOfs;
} // ofs最大值为maxOfs
if (ofs > maxOfs)
ofs = maxOfs; // Make offsets relative to base
// 之前的ofs和lastOfs都是相对于hint位置的， 现在把它重置为相对于base的位置
lastOfs += hint;
ofs += hint;
} else { // key <= a[base + hint]
// 从base+hint向前搜索，查找区间[base, base + hint]， 直到找到ofs值使得a[base+hint-ofs] < key <= a[base+hint-lastOfs]
// maxOfs为hint+1， 而ofs < maxOfs, 因此当ofs = maxOfs -1 时， 比较到的最左边的元素为a[base + hint - hint] == a[base]
final int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
lastOfs = ofs;
// ofs乘以2再加1
ofs = (ofs << 1) + 1;
// 正整数溢出
if (ofs <= 0)
ofs = maxOfs;
}
// 最大为maxOfs
if (ofs > maxOfs)
ofs = maxOfs; // 重置ofs和lastOfs为相对于base的位置索引
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; /*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
* to the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
*/
lastOfs++;
// 查找准确位置
while (lastOfs < ofs) {
// 中间位置
int m = lastOfs + ((ofs - lastOfs) >>> 1); if (c.compare(key, a[base + m]) > 0)
lastOfs = m + 1; // a[base + m] < key
else
ofs = m; // key <= a[base + m]
}
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
} /**
* Like gallopLeft, except that if the range contains an element equal to
* key, gallopRight returns the index after the rightmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
private static <T> int gallopRight(T key, T[] a, int base, int len,
int hint, Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len; int ofs = 1;
int lastOfs = 0;
if (c.compare(key, a[base + hint]) < 0) {
// 从base + hint位置向前搜索区间[base, base + hint]， 使得a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
// 记录上次查找位置
lastOfs = ofs;
// 乘以2 加1
ofs = (ofs << 1) + 1;
// 正整数溢出
if (ofs <= 0)
ofs = maxOfs;
}
// 最大为maxOfs
if (ofs > maxOfs)
ofs = maxOfs; // 重置ofs和lastOfs为相对于base的位置索引
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
} else { // a[b + hint] <= key
// 搜索区间[base + hint, base + len -1]使得a[b+hint + lastOfs] <= key < a[b+hint + ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
// 正整数溢出
if (ofs <= 0)
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs; // 重置ofs和lastOfs为相对于base的位置索引
lastOfs += hint;
ofs += hint;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; lastOfs++;
// 查找key的准确位置
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1); if (c.compare(key, a[base + m]) < 0)
ofs = m; // key < a[b + m]
else
lastOfs = m + 1; // a[b + m] <= key
}
// 最终会找到m使得k >= a[base + m]， 而此时lastOfs == ofs且lastOfs = m +1， 则ofs = m +1， 因此a[base + ofs] > k >= a[base + ofs -1]， ofs即m+1， ofs - 1即为m， 因此ofs位置的值大于key
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
} /**
*基于以上gallopRight方法最后查找key的索引的解释， 因此run1的第一个元素一定大于run2的第一个元素， *而run2中run1最后一个元素所在索引位置之后的值也被忽略掉， 因此run1的最后一个元素大于run2中的所有元素的值。*/ /**
* Merges two adjacent runs in place, in a stable fashion. The first
* element of the first run must be greater than the first element of the
* second run (a[base1] > a[base2]), and the last element of the first run
* (a[base1 + len1-1]) must be greater than all elements of the second run.
*
* For performance, this method should be called only when len1 <= len2;
* its twin, mergeHi should be called if len1 >= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2; // Copy first run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len1);
int cursor1 = tmpBase; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
// 目标位置从base1的索引开始， 因为base1在base2之前， 下面会将base1的内容放入临时数组， 这样run1中的内容就可以覆盖了
int dest = base1; // Indexes int a
// 把第一个序列的内容放入临时数组tmp中
System.arraycopy(a, base1, tmp, cursor1, len1); // Move first element of second run and deal with degenerate cases
// 因为run1的第一个元素大于run2的第一个元素， 因此将run2的第一个元素先放进
a[dest++] = a[cursor2++];
// 如果run2只有一个元素， 则将run1中剩余的元素放入正确位置后返回
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
return;
}
// 如果run1只有一个元素， 因为run1的最后一个元素大于run2中的所有元素， 因此先将run2中的元素放入正确位置， 然后将run1的唯一的一个元素放入最后一个位置， 然后返回
if (len1 == 1) {
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
return;
} Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won /*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
assert len1 > 1 && len2 > 0;
if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
a[dest++] = a[cursor2++];
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
a[dest++] = tmp[cursor1++];
count1++;
count2 = 0;
if (--len1 == 1)
break outer;
}
// 当每个序列中的连续放入目标位置的元素个数小于minGallop时， 这样分别拷贝就可以了
} while ((count1 | count2) < minGallop); /*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
// 因为两个run序列的大小是近似相等的， 如果一个序列连续超过minGallop个数的元素被放入目标位置， 则另一个有接近大小的连续序列等待被放入正确位置，切换成Gallopping模式
do {
assert len1 > 1 && len2 > 0;
//查找run1第一个大于run2中第一个元素的元素的位置索引
count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
if (count1 != 0) {
// run1中count1之前的元素全部放入目标序列
System.arraycopy(tmp, cursor1, a, dest, count1);
// 移动索引位置
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
// 移动run2的第一个元素到目标序列中
a[dest++] = a[cursor2++];
// 如果run2中没有其他元素则跳出
if (--len2 == 0)
break outer; // 查找run2中第一个小于等于run1当前元素的元素的位置索引
count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
if (count2 != 0) {
// 拷贝count2之后的元素到目标序列
System.arraycopy(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
// run2中没有其他元素则跳出
if (len2 == 0)
break outer;
} // 此时run2中的第一个元素大于等于run1中的第一个元素， 拷贝run1中的第一个元素到目标序列
a[dest++] = tmp[cursor1++];
// 如果run1中只有一个元素则跳出
if (--len1 == 1)
break outer;
// 动态调整minGallop的值
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
// 调整minGallop的值， 使得在有序序列不多的情况下不用Gallopping模式
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field // run1中只有一个元素
if (len1 == 1) {
assert len2 > 0;
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
} else if (len1 == 0) {
// 因为run1中的最后一个元素大于run2中的所有元素， 因此这种情况不存在
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
// run2中已经没有元素
assert len2 == 0;
assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
} /**
* Like mergeLo, except that this method should be called only if
* len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
private void mergeHi(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2; // Copy second run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len2);
int tmpBase = this.tmpBase;
// 将run2中的所有元素放入临时数组tmp中
System.arraycopy(a, base2, tmp, tmpBase, len2); int cursor1 = base1 + len1 - 1; // Indexes into a
int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array
// 从后往前插入元素
int dest = base2 + len2 - 1; // Indexes into a // Move last element of first run and deal with degenerate cases
// run1的最后一个元素导入目标位置
a[dest--] = a[cursor1--];
// 如果run1中只有一个元素， 将run2中的剩余元素放入目标位置（从后往前）
if (--len1 == 0) {
System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
return;
}
if (len2 == 1) {
// run2中只有一个元素， 因为run1的第一个元素大于run2的第一个元素， 因此， run2中唯一的一个元素小于run1中所有的元素， 因此将run1中的元素全部放入目标位置， 最后将唯一的run2中的一个元素放入第一个位置
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2];
return;
} Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won /*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
assert len1 > 0 && len2 > 1;
if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
// 从后往前放入目标位置
a[dest--] = a[cursor1--];
count1++;
count2 = 0;
// run1中没有了元素
if (--len1 == 0)
break outer;
} else {
a[dest--] = tmp[cursor2--];
count2++;
count1 = 0;
// run2中只有一个剩余元素
if (--len2 == 1)
break outer;
}
} while ((count1 | count2) < minGallop); /*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 0 && len2 > 1;
// 找到大于run2当前位置的元素的run1中元素， 因为是从后往前查找， 因此找到的位置比如k1，k1之后的所有元素大于run1, run2中的剩余所有元素
count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
if (count1 != 0) {
dest -= count1;
cursor1 -= count1;
len1 -= count1;
// 拷贝run1中的大的元素
System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
// run1中没有元素， 跳出
if (len1 == 0)
break outer;
}
// run2的当前元素拷贝到目标位置
a[dest--] = tmp[cursor2--];
if (--len2 == 1)
break outer; // 找到run2中大于等于run1当前元素的元素的位置索引比如K2， 则k2之后的所有元素大于run1, run2中的剩余元素
count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1) // len2 == 1 || len2 == 0
break outer;
}
// 拷贝run1的当前元素到目标位置， 因为a[cursior1]大于等于run2中的剩余元素
a[dest--] = a[cursor1--];
if (--len1 == 0)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field // run2中只有一个元素
if (len2 == 1) {
assert len1 > 0;
dest -= len1;
cursor1 -= len1;
// 拷贝run1中的所有元素到目标位置
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
// run2的最后一个元素放入第一个位置
a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
} else if (len2 == 0) {
// 因为run2的第一个元素小于run1， run2中的所有元素， 因此run2不可能最后为空
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len1 == 0;
assert len2 > 0;
// 拷贝run2中的剩余元素
System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
}
} /**
* Ensures that the external array tmp has at least the specified
* number of elements, increasing its size if necessary. The size
* increases exponentially to ensure amortized linear time complexity.
*
* @param minCapacity the minimum required capacity of the tmp array
* @return tmp, whether or not it grew
*/
private T[] ensureCapacity(int minCapacity) {
if (tmpLen < minCapacity) {
// Compute smallest power of 2 > minCapacity
int newSize = minCapacity;
newSize |= newSize >> 1;
newSize |= newSize >> 2;
newSize |= newSize >> 4;
newSize |= newSize >> 8;
newSize |= newSize >> 16;
newSize++; if (newSize < 0) // Not bloody likely!
newSize = minCapacity;
else
newSize = Math.min(newSize, a.length >>> 1); @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
T[] newArray = (T[])java.lang.reflect.Array.newInstance
(a.getClass().getComponentType(), newSize);
tmp = newArray;
tmpLen = newSize;
tmpBase = 0;
}
return tmp;
}
}
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