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：

 class Solution {
 public:
     vector<int> searchRange(vector<int>& nums, int target) {
         int sz = nums.size();
         int left = bs(nums, , sz, target, true);
         int right = bs(nums, , sz, target, false);
         vector<int> res;
         res.push_back(left);
         res.push_back(right);
         return res;
     }

     int bs(vector<int>& nums, int beg, int end, int target, int goLeft)
     {
         if(beg > end)
             return -;
         int mid = (beg + end)/;
         if(nums[mid] == target){
             int tmpAns = (goLeft == true ? bs(nums, beg, mid - , target, goLeft) : bs(nums, mid + , end, target, goLeft));
             return tmpAns == - ? mid : tmpAns;
         }else if(nums[mid] < target){
             return bs(nums, mid + , end, target, goLeft);
         }else{
             return bs(nums, beg, mid - , target, goLeft);
         }
     }
 };