34. Search for a Range 史上最简洁的二分搜索的写法

Given an array of integers sorted in ascending order, 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].

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int n = nums.size(), L = 0, R = n - 1;        if(n == 0) return {-1, -1};        while(L < R){            int mid = (L + R) / 2;// 在L<R的情况下，因为每次二分后的mid必小于R大于或等于L，由L = mid + 1;R = mid;可知：每经过一次循环，L与R的距离必定-1。最终的结果一定是L == R，那么检查nums[L] == target是否成立即可。            if(nums[mid] < target) L = mid + 1;            else R = mid;        }        if(nums[L] != target) return {-1, -1};        int left = L, right = L;        while(nums[left] == nums[left - 1] && left > 0) left--;        while(nums[right] == nums[right + 1] && right < n -1) right++;        return {left, right};            }};