63.排序数组中找元素的第一个元素和最后一个元素

苦藤新鸡2026-02-20 8:48

1.题目描述

示例 1：

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输入：nums = [5,7,7,8,8,10], target = 8
输出：[3,4]

示例 2：

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输入：nums = [5,7,7,8,8,10], target = 6
输出：[-1,-1]

示例 3：

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输入：nums = [], target = 0
输出：[-1,-1]

2.代码

cpp 复制代码

/**
 * 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;
}