开发公司硬底化路面工程入账长沙官网seo收费标准

1. 题目

2. 思路和题解

这道题虽然是道中等题，并且看起来很复杂，但是实际上就是给定一个数组和目标值，让我们去寻找该目标值在数组中的位置。题目还提到说设计O(log n)的算法解决问题，更进一步暗示我们去用二分查找。要找开始位置和结束位置，那就分两步：

1. 寻找开始位置

while (left <= right) {int mid = left + (right - left) / 2;if (nums[mid] == target) {arr[0] = mid; //找到目标值后，将下标赋值给数组第一个元素right = mid - 1; // 这一步很重要，是为了寻找在这之前，是否还存在目标值} else if (nums[mid] < target){left = mid + 1;} else {right = mid - 1;}
}

2. 寻找结束位置

while (left <= right) {int mid = left + (right - left) / 2;if (nums[mid] == target) {arr[1] = mid;left = mid + 1; //这一步和上面一样也很重要，是为了确定在这之后，是否还存在目标值} else if (nums[mid] < target){left = mid + 1;} else {right = mid - 1;}
}

所以整体代码如下

class Solution {public int[] searchRange(int[] nums, int target) {int left = 0;int right = nums.length - 1;int[] arr = {-1, -1};//第一次查找while (left <= right) {int mid = left + (right - left) / 2;if (nums[mid] == target) {arr[0] = mid;right = mid - 1;} else if (nums[mid] < target){left = mid + 1;} else {right = mid - 1;}}//第二次查找left = 0;right = nums.length - 1;while (left <= right) {int mid = left + (right - left) / 2;if (nums[mid] == target) {arr[1] = mid;left = mid + 1;} else if (nums[mid] < target){left = mid + 1;} else {right = mid - 1;}}return arr;}
}