C语言刷数组题记录
2024-10-08 01:56:49
讲解：https://mp.weixin.qq.com/s/weyitJcVHBgFtSc19cbPdw
二分法：
704. 二分查找
int search(int* nums, int numsSize, int target)
{
int left = 0;
int right = numsSize;
while (left < right) {
// int cur = (left + right) / 2;
int cur = left + (right - left) / 2;
if (nums[cur] == target) {
return cur;
} else if (nums[cur] > target) {
right = cur;
} else {
left = cur + 1;
}
}
return -1;
}
27. 移除元素
暴力法（两层for循环）和 双指针
int removeElement(int* nums, int numsSize, int val)
{
for (int i = 0; i < numsSize; i++) {
if (nums[i] == val) {
// 数组整体向前移一步
for (int j = i; j < numsSize - 1; j++) {
nums[j] = nums[j + 1];
}
numsSize--;
i--;
}
}
return numsSize;
}
双指针（快慢指针）
int removeElement(int* nums, int numsSize, int val)
{
int slow = 0;
for (int fast = 0; fast < numsSize; fast++) {
if (nums[fast] != val) {
nums[slow++] = nums[fast];
}
}
return slow;
}
977. 有序数组的平方
/**
* Note: The returned array must be malloced, assume caller calls free().
*/ int Double(int a)
{
return a * a;
} int* sortedSquares(int* nums, int numsSize, int* returnSize)
{
int left = 0;
int right = numsSize - 1;
int index = right;
*returnSize = numsSize;
int *res = (int *)malloc(sizeof(int) * numsSize); while (left <= right) {
if (Double(nums[left]) <= Double(nums[right])) {
printf("1 %d <= %d index=%d\n", Double(nums[left]), Double(nums[right]), index);
res[index--] = Double(nums[right]);
right--;
} else {
printf("2 %d > %d index=%d\n", Double(nums[left]), Double(nums[right]), index);
res[index--] = Double(nums[left]);
left++;
}
}
return res;
}
209. 长度最小的子数组
滑动窗口
int minSubArrayLen(int target, int* nums, int numsSize)
{
int slow = 0;
int fast = 0;
int sum = 0;
int res = 1e7;
for (int fast = 0; fast < numsSize; fast++) {
sum += nums[fast];
while (sum >= target) {
int len = fast - slow + 1;
res = res < len ? res : len;
// sum = 0; 这是不对的！
printf("slow:%d\n", slow);
sum -= nums[slow++];
}
} return res == 1e7 ? 0 : res;
}
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