Search for a Range--LeetCode

1.题目

Search for a Range

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].

2.题意

给定升序数组，找到target的起始和结束位置
若不存在，返回 [-1,-1]
算法的复杂度必须是 O (log n)

3.分析

借助两次二分查找即可找出左右边界

4.代码

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> result(2, -1);        if(nums.size() == 0)            return result;        int ll = 0;        int lr = nums.size() - 1;        while(ll <= lr)        {            int m = ll + (lr - ll) / 2;            if(nums[m] < target)                ll = m + 1;            else                lr = m - 1;        }        int rl = 0;        int rr = nums.size() - 1;        while(rl <= rr)        {            int m = rl + (rr - rl) / 2;            if(nums[m] > target)                rr = m - 1;            else                rl = m + 1;        }        if(ll <= rr)        {            result[0] = ll;            result[1] = rr;        }        return result;    }};