LeetCode34 Search for a Range

题目：

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].  （Medium）

分析：

标准的二分搜索题，找一个target的范围，也就是first position 元素等于target 和last position元素等于。

写两次二分搜索，注意中间start，end的变化情况的不同，一个为了保留住第一个满足条件的，一个为了保留住最后一个满足条件的。

代码：

 class Solution {
 public:
     vector<int> searchRange(vector<int>& nums, int target) {
         vector<int> result;
         if (nums.size() == ) {
             return result;
         }
         int start = , end = nums.size() - ;
         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (nums[mid] < target) {
                 start = mid;
             }
             else {
                 end = mid;
             }
         }
         if (nums[start] == target) {
             result.push_back(start);
         }
         else if (nums[end] == target) {
             result.push_back(end);
         }

         start = ;
         end = nums.size() - ;
         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (nums[mid] <= target) {
                 start = mid;
             }
             else {
                 end = mid;
             }
         }
         if (nums[end] == target) {
             result.push_back(end);
         }
         else if (nums[start] == target) {
             result.push_back(start);
         }
         if (result.size() != ) {
             return result;
         }
         else {
             return vector<int> {-, -};
         }
     }
 };