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：

 // LeetCode, Search in Rotated Sorted Array II
 // 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     bool search(vector<int>&nums, int target) {
         int first = , last = nums.size()-;
         while (first <= last) {
             const int mid = (first + last) / ;
             if (nums[mid] == target)
                 return true;
             if (nums[first] < nums[mid]) {
                 if (nums[first] <= target && target < nums[mid])
                     last = mid - ;
                 else
                     first = mid + ;
             } else if (nums[first] > nums[mid]) {
                 if (nums[mid] < target && target <= nums[last])
                     first = mid + ;
                 else
                     last = mid;
             } else
                 //skip duplicate one
                 first++;
         }
         return false;
     }
 };