LeetCode之链表

2. Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

暴力解法：

解题思路：

1.其中一个链表为空的情况；

2.链表长度相同时，且最后一个节点相加有进位的情况；

3.链表长度不等，短链表最后一位有进位的情况，且进位后长链表也有进位的情况；

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) { val = x; }

 * }

 */

public class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        if(l1==null){

            return l2;

        }

        if(l2==null){

            return l1;

        }

        int len1 = getLength(l1);

        int len2 = getLength(l2);

        ListNode head = null;

        ListNode plus = null;

        if(len1>=len2){

            head = l1;

            plus = l2;

        }else{

            head = l2;

            plus = l1;

        }

        ListNode p = head;

        int carry = 0;

        int sum = 0;

        while(plus!=null){

            sum = p.val + plus.val+carry;

            carry = sum/10;

            p.val = sum%10;

            if(p.next==null&&carry!=0){

                ListNode node = new ListNode(carry);

                p.next = node;

                carry = 0;

            }

            p = p.next;

            plus = plus.next;

        }

        while(p!=null&&carry!=0){

            sum = p.val+carry;

            carry = sum/10;

            p.val = sum%10;

            if(p.next==null&&carry!=0){

                ListNode node = new ListNode(carry);

                p.next = node;

                carry=0;

            }

            p = p.next;

        }

        return head;

    }

    public int getLength(ListNode l){

        if(l==null){

            return 0;

        }

        int len = 0;

        while(l!=null){

            ++len;

            l = l.next;

        }

        return len;

    }

}

递归解法：

复杂度

时间O(n) 空间(n) 递归栈空间

思路

从末尾到首位，对每一位对齐相加即可。技巧在于如何处理不同长度的数字，以及进位和最高位的判断。这里对于不同长度的数字，我们通过将较短的数字补0来保证每一位都能相加。递归写法的思路比较直接，即判断该轮递归中两个ListNode是否为null。

全部为null时，返回进位值
有一个为null时，返回不为null的那个ListNode和进位相加的值
都不为null时，返回两个ListNode和进位相加的值

public class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        return helper(l1,l2,0);

    }

    public ListNode helper(ListNode l1, ListNode l2, int carry){

        if(l1==null && l2==null){

            return carry == 0? null : new ListNode(carry);

        }

        if(l1==null && l2!=null){

            l1 = new ListNode(0);

        }

        if(l2==null && l1!=null){

            l2 = new ListNode(0);

        }

        int sum = l1.val + l2.val + carry;

        ListNode curr = new ListNode(sum % 10);

        curr.next = helper(l1.next, l2.next, sum / 10);

        return curr;

    }

}

迭代法：

复杂度

时间O(n) 空间(1)

思路

迭代写法相比之下更为晦涩，因为需要处理的分支较多，边界条件的组合比较复杂。过程同样是对齐相加，不足位补0。迭代终止条件是两个ListNode都为null。

注意

迭代方法操作链表的时候要记得手动更新链表的指针到next
迭代方法操作链表时可以使用一个dummy的头指针简化操作
不可以在其中一个链表结束后直接将另一个链表串接至结果中，因为可能产生连锁进位

public class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        ListNode dummyHead = new ListNode(0);

        if(l1 == null && l2 == null){

            return dummyHead;

        }

        int sum = 0, carry = 0;

        ListNode curr = dummyHead;

        while(l1!=null || l2!=null){

            int num1 = l1 == null? 0 : l1.val;

            int num2 = l2 == null? 0 : l2.val;

            sum = num1 + num2 + carry;

            curr.next = new ListNode(sum % 10);

            curr = curr.next;

            carry = sum / 10;

            l1 = l1 == null? null : l1.next;

            l2 = l2 == null? null : l2.next;

        }

        if(carry!=0){

            curr.next = new ListNode(carry);

        }

        return dummyHead.next;

    }

}

21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

思路1:1）考虑特殊情况，两个链表至少有一个为空；

　　 2）考虑两个链表长度不一样的情况；

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) { val = x; }

 * }

 */

public class Solution {

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

        if(l1==null)

            return l2;

        if(l2==null)

            return l1;

        ListNode head;

        if(l1.val<=l2.val){

            head = new ListNode(l1.val);

            l1 = l1.next;

        }else{

            head = new ListNode(l2.val);

            l2 = l2.next;

        }

        ListNode node = head;

        while(l1!=null&&l2!=null){

            if(l1.val<=l2.val){

                node.next = l1;

                node = node.next;

                l1 = l1.next;

            }else{

                node.next = l2;

                node = node.next;

                l2 = l2.next;

            }

        }

        if(l1!=null){

            node.next = l1;

        }

        if(l2!=null){

            node.next = l2;

        }

        return head;

    }

}

206. Reverse Linked List

Reverse a singly linked list.

思路：1）首先判断头节点是否为空，为空则直接返回；

　　2）

public class Solution {

    public ListNode reverseList(ListNode head) {

        if(head==null)

            return null;

        ListNode newHead = new ListNode(head.val);

        ListNode p = head.next;

        while(p!=null){

            ListNode node = new ListNode(p.val);

            node.next = newHead;

            p = p.next;

            newHead = node;

        }

        return newHead;

    }

}

sort-list

Sort a linked list in O(n log n) time using constant space complexity.

思路：要求时间复杂度为O(nlogn)，空间复杂度为O(1)；

　　1.考虑使用归并排序；

　　2.归并排序需要找到中点，考虑使用快慢指针；

　　3.归并中排序；

/**

 * Definition for singly-linked list.

 * class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) {

 *         val = x;

 *         next = null;

 *     }

 * }

 */

public class Solution {

    public ListNode sortList(ListNode head) {

        if(head==null||head.next==null)

            return head;

        ListNode mid = getMid(head);

        ListNode right = sortList(mid.next);

        mid.next = null;

        ListNode left = sortList(head);

        return mergeList(left,right);

    }

    public ListNode getMid(ListNode head){

        ListNode slow = head;

        ListNode fast = head.next;

        while(fast!=null&&fast.next!=null){

            slow = slow.next;

            fast = fast.next.next;

        }

        return slow;

    }

    public ListNode mergeList(ListNode left,ListNode right){

        if(left==null)

            return right;

        if(right==null)

            return left;

        ListNode head = null;

        if(left.val<=right.val){

            head = left;

            left = left.next;

        }else{

            head = right;

            right = right.next;

        }

        ListNode node = head;

        while(left!=null&&right!=null){

            if(left.val<=right.val){

                node.next = left;

                left = left.next;

            }else{

                node.next = right;

                right = right.next;

            }

            node = node.next;

        }

        if(left!=null){

            node.next = left;

        }

        if(right!=null){

            node.next = right;

        }

        return head;

    }

}

insertion-sort-list

Sort a linked list using insertion sort.

使用插入的方式对链表进行排序：

插入排序的思路如下：当前数与其前面的有序数依次进行比较，找到适当的位置插入，以此类推，得到最终结果；

此题的思路是：

　　1）如果head为空或者head.next为空，则直接返回head;

　　2)使用一个节点cur遍历当前待排序的节点，利用另一个节点pre从头开始遍历，若pre的值<=cur的值且两者不同，pre=pre.next；否则，记录第一个>cur的节点及其值，再遍历此节点到cur节点，调整节点值得位置，将cur的值交换到第一个>cur的节点处，直到链表遍历完；

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) {

 *         val = x;

 *         next = null;

 *     }

 * }

 */

public class Solution {

    public ListNode insertionSortList(ListNode head) {

        if(head==null||head.next==null){

            return head;

        }

        ListNode cur = head;

        while(cur!=null){

            ListNode pre = head;

            while(pre.val<=cur.val&&pre!=cur){

                pre = pre.next;

            }

            int firstVal = pre.val;

            ListNode mark = pre;

            while(pre!=cur){

                int nextVal = pre.next.val;

                int tmp = nextVal;

                pre.next.val = firstVal;

                firstVal = tmp;

                pre = pre.next;

            }

            mark.val = firstVal;

            cur = cur.next;

        }

        return head;

    }

}

reorder-list

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.

解题思路：

　　先使用快慢指针找到链表的中点，反转后半部分的链表，再以中点为断点进行前后两部分交叉合并；

/**

 * Definition for singly-linked list.

 * class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) {

 *         val = x;

 *         next = null;

 *     }

 * }

 */

public class Solution {

    public void reorderList(ListNode head) {

        if(head==null||head.next==null)

            return;

        ListNode fast = head.next;

        ListNode slow = head;

        while(fast!=null&&fast.next!=null){

            slow = slow.next;

            fast = fast.next.next;

        }

        ListNode pre = reverseList(slow.next);

        slow.next = pre;

        ListNode p = head;

        ListNode q = slow.next;

        while(q!=null&&p!=null){

            slow.next = q.next;

            q.next = p.next;

            p.next = q;

            p = q.next;

            q = slow.next;

        }

    }
　　//反转单链表

    public ListNode reverseList(ListNode head){

        if(head==null||head.next==null)

            return head;

        ListNode cur = head.next;

        ListNode pre = head;

        while(cur!=null){

            ListNode node = cur.next;

            cur.next = pre;

            pre = cur;

            cur = node;

        }

        head.next = cur;

        return pre;

    }

}

巴特西

LeetCode之链表

2. Add Two Numbers

复杂度

思路

注意

21. Merge Two Sorted Lists

206. Reverse Linked List

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