34. Search for a Range(unsolved)

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) {        vector<int> res(2,-1);        if(nums.empty()) return res;        int left=0,right=nums.size()-1;        while(left<right)        {            int mid=(left+right)/2;            if(nums[mid]<target) left=mid+1;            else right=mid;        }        if(nums[right]!=target) return res;        res[0]=right;        left=0;right=nums.size();        while(left<right)        {            int mid=(left+right)/2;             if(nums[mid]<=target) left=mid+1;            else right=mid;        }        res[1]=left-1;        return res;    }};

解答：这道题就是先用二分找左值，然后二分找右值。
要记住套路，找左值就是

int left=0,right=nums.size()-1;        while(left<right)        {            int mid=(left+right)/2;            if(nums[mid]<target) left=mid+1;            else right=mid;        } res[0]=right;

找右值就是

left=0;right=nums.size();        while(left<right)        {            int mid=(left+right)/2;             if(nums[mid]<=target) left=mid+1;            else right=mid;        }        res[1]=left-1;