63.排序数组中找元素的第一个元素和最后一个元素
1.题目描述
示例 1：
输入：nums = [5,7,7,8,8,10], target = 8输出：[3,4]示例 2：
输入：nums = [5,7,7,8,8,10], target = 6输出：[-1,-1]示例 3：
输入：nums = [], target = 0输出：[-1,-1]
2.代码
/** * Note: The returned array must be malloced, assume caller calls free(). */ int find_left_bound(int* nums, int numsSize, int target){ int left = 0; int right = numsSize - 1; int flag = 0; while(left <= right){ int mid = left + (right - left )/2; if(nums[mid] == target){ right = mid - 1;//找到了就往左边找左边界 flag = 1; } if(nums[mid] > target){ right = mid - 1; } if(nums[mid] < target){ left = mid + 1; } } //为什么会返回left呢？ //当left == right的时候最后一次循环，如何right左移动，left才指向target if(flag == 0){ return -1; } return left; } int find_right_bound(int* nums, int numSize, int target){ int left = 0; int right = numSize - 1; int flag = 0; while(left <= right){ int mid = left + (right - left )/2; if(nums[mid] == target){ left = mid + 1;//找到了就往左边找左边界 flag = 1; } if(nums[mid] > target){ right = mid - 1; } if(nums[mid] < target){ left = mid + 1; } } if(flag == 0){ return -1; } //此时返回right return right; } int* searchRange(int* nums, int numsSize, int target, int* returnSize) { int left_bound = find_left_bound(nums,numsSize,target); int right_bound = find_right_bound(nums,numsSize,target); int* res = (int*)malloc(sizeof(int)*2); res[0] = left_bound; res[1] = right_bound; returnSize[0] = 2; return res; }