LeetCode OJ：Search in Rotated Sorted Array II（翻转排序数组的查找）

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

这个和前面的一个不一样就在于可能会有重复的数字，那么判断的时候就应该注意了，遇到start<=mid的不一定说明当前的区间[start, mid]就是递增的。这时候就应该++来确认下，代码如下：

 1 // LeetCode, Search in Rotated Sorted Array II
 2 // 时间复杂度O(n)，空间复杂度O(1)
 3 class Solution {
 4 public:
 5     bool search(vector<int>&nums, int target) {
 6         int first = 0, last = nums.size()-1;
 7         while (first <= last) {
 8             const int mid = (first + last) / 2;
 9             if (nums[mid] == target)
10                 return true;
11             if (nums[first] < nums[mid]) {
12                 if (nums[first] <= target && target < nums[mid])
13                     last = mid - 1;
14                 else
15                     first = mid + 1;
16             } else if (nums[first] > nums[mid]) {
17                 if (nums[mid] < target && target <= nums[last])
18                     first = mid + 1;
19                 else
20                     last = mid;
21             } else
22                 //skip duplicate one
23                 first++;
24         }
25         return false;
26     }
27 };