LeetCode OJ：Search for a Range（区间查找）

Given a sorted array of integers, 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].

查找一个区间开头以及结尾的下标，分两次二分查找，一次向左一次向右即可，代码如下：

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         int sz = nums.size();
 5         int left = bs(nums, 0, sz, target, true);
 6         int right = bs(nums, 0, sz, target, false);
 7         vector<int> res;
 8         res.push_back(left);
 9         res.push_back(right);
10         return res;
11     }
12
13     int bs(vector<int>& nums, int beg, int end, int target, int goLeft)
14     {
15         if(beg > end)
16             return -1;
17         int mid = (beg + end)/2;
18         if(nums[mid] == target){
19             int tmpAns = (goLeft == true ? bs(nums, beg, mid - 1, target, goLeft) : bs(nums, mid + 1, end, target, goLeft));
20             return tmpAns == -1 ? mid : tmpAns;
21         }else if(nums[mid] < target){
22             return bs(nums, mid + 1, end, target, goLeft);
23         }else{
24             return bs(nums, beg, mid - 1, target, goLeft);
25         }
26     }
27 };