leetcode week11

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].

问题描述：给定从小到大的数组，找出给定值的位置。要求复杂度是O(log n)，可以看出要用二分法。

解题思路：利用二分法，先找出target所在的位置。由于题目要求的是找出target的范围，因此找出之后再判断两边是不是一样的数，直到找出最左target与最右target的index。

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