[leetcode] 34. Search for a Range 解题报告

题目链接：https://leetcode.com/problems/search-for-a-range/

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

思路：使用二分查找找到左右边界即可，找到左右边界只要改变一下判断条件即可。

如果搜索条件是: if(nums[mid] <= target) left = mid+1; 这样left最终会停在结果的右边一位

如果搜索条件是: if(nums[mid] < target) left = mid+1; 这样left最终会停在结果的第一个位置

这样如果两个条件下搜索结果left停留在相同的位置, 说明没有搜索到, 就可以返回默认的[-1, -1]了, 否则就返回正确的位置.

代码如下：

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        if(nums.size()==0) return {-1, -1};        int len = nums.size(), left = 0, right = len-1, ans;        while(left <= right)        {            int mid = (left+right)/2;            if(nums[mid] >= target) right = mid-1;            else left = mid+1;        }        ans = left, right = len-1;        while(left <= right)        {            int mid = (left+right)/2;            if(nums[mid] > target) right = mid-1;            else left = mid+1;        }        if(left==ans) return {-1, -1};        return {ans, left-1};    }};