81. Search in Rotated Sorted Array II

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

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

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

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

The array may contain duplicates.

和33题一样，但是可以包含duplicate，那么就是要除了判断排好的情况，还要判断一下，头尾可能和mid相同的情况，这样就直接start++ 或者 end--

代码如下：

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        if (nums.size() == 0) return false;
        int start = 0, end = nums.size() - 1;
        int mid;
        while (start <= end) {
            mid = (start + end) / 2;
            if (nums[mid] == target) return true;
            
            if (nums[mid] < nums[end]) {
                if (nums[mid] < target && target <= nums[end])
                    start = mid + 1;
                else end = mid - 1;
            }
            else if (nums[mid] > nums[start]) {
                if (nums[mid] > target && target >= nums[start])
                    end = mid - 1;
                else start = mid + 1;
            }
            else {//说明头尾必定有一个和mid的值相同
                if (nums[mid] == nums[start])
                    start++;
                else end--;
            }
        }
        return false;
    }
};