leetcode x的平方根 c语言

题目描述：求解LeetCode平方根。

解题思路：可以使用二分查找法，在区间[0, x]内查找平方根。每次取中间值mid，如果mid的平方小于x，则在[mid, x]中继续查找；如果mid的平方大于x，则在[0, mid]中继续查找；如果mid的平方等于x，则返回mid。这样直到区间缩小到一个足够小的范围，即可得到近似的平方根。

以下是Python代码实现：

class Solution:
    def mySqrt(self, x: int) -> int:
        l, r = 0, x
        while l <= r:
            mid = (l + r) // 2
            if mid * mid <= x < (mid + 1) * (mid + 1):
                return mid
            elif x < mid * mid:
                r = mid - 1
            else:
                l = mid + 1

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leetcode c语言

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leetcode 第102题 c语言实现

可以使用队列来实现二叉树的层序遍历，具体实现可以参考以下代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes){
    if (root == NULL) {
        *returnSize = ;
        return NULL;
    }
    int** res = (int**)malloc(sizeof(int*) * 100);
    *returnColumnSizes = (int*)malloc(sizeof(int) * 100);
    int front = , rear = ;
    struct TreeNode* queue[100];
    queue[rear++] = root;
    while (front < rear) {
        int size = rear - front;
        (*returnColumnSizes)[*returnSize] = size;
        res[*returnSize] = (int*)malloc(sizeof(int) * size);
        for (int i = ; i < size; i++) {
            struct TreeNode* node = queue[front++];
            res[*returnSize][i] = node->val;
            if (node->left) {
                queue[rear++] = node->left;
            }
            if (node->right) {
                queue[rear++] = node->right;
            }
        }
        (*returnSize)++;
    }
    return res;
}