【Leetcode】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?

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

class Solution {
public:
    bool search(int A[], int n, int target) {
        int start = 0, end = n - 1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (A[mid] == target) return true;
            if (A[start] < A[mid]) {
                if (A[start] <= target && A[mid] > target) {
                    end = mid - 1;
                } else {
                    start = mid + 1;
                }
            } else if (A[start] > A[mid]) {
                if (A[mid] < target && A[end] >= target) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            } else {
                ++start;
            }
        }
        return false;
    }
};

整体上与上题（Search in Rotated Sorted Array）差别不大，额外的是：若数组中存在重复元素，那么区间首尾元素相等时无法确定区间内是否发生过rotate，此时跳过首元素继续找。