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 = 0, last = nums.size()-1;
        while (first <= last) {
            const int mid = (first + last) / 2;
            if (nums[mid] == target)
                return true;
            if (nums[first] < nums[mid]) {
                if (nums[first] <= target && target < nums[mid])
                    last = mid - 1;
                else
                    first = mid + 1;
            } else if (nums[first] > nums[mid]) {
                if (nums[mid] < target && target <= nums[last])
                    first = mid + 1;
                else
                    last = mid;
            } else
                //skip duplicate one
                first++;
        }
        return false;
    }
};