[leetcode]34. Search for a Range

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

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].

方法一：

class Solution{public:    vector<int> searchRange(vector<int>& nums,int target)    {        vector<int> res(2,-1);        if(nums.empty())            return res;        int i=0,j=nums.size()-1;        //找到左边起始位置        while(i<j)        {            int mid=i+(j-i)/2;            if(nums[mid]<target)                i=mid+1;            else                j=mid;        }        if(nums[i]!=target)            return res;        else            res[0]=i;        //找到右边结束位置        j=nums.size()-1;        while(i<j)        {            int mid=i+(j-i)/2+1;            if(nums[mid]>target)                j=mid-1;            else                i=mid;        }        res[1]=j;        return res;    }};

方法二：

class Solution{public:    vector<int> searchRange(vector<int>& nums,int target)    {        auto bounds=equal_range(nums.begin(),nums.end(),target);        if(bounds.first==bounds.second)            return {-1,-1};        return {bounds.first-nums.begin(),bounds.second-nums.begin()-1};    }};