Search in Rotated Sorted Array II

题目：

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.

分析：

复杂度是肯定要变差的，如果所有元素都是1，就一个是2，然后找2，二分法的特性就被破坏了，所以worst case变成linear的了。

解法：

class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
public:
    bool search(vector<int> &A, int target) {
        // write your code here
        if (A.size() == 0) {
            return false;
        }
        int start = 0;
        int end = A.size() - 1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return true;
            }

            if (A[start] == A[end] || A[mid] == A[end]) {
                end--;
                continue;
            }

            if (A[mid] < A[end]) {
                if (A[mid] < target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            } else {
                if (target < A[mid] && A[start] <= target) {
                    end = mid;
                } else {
                    start = mid;
                }
            }
        }

        if (A[start] == target || A[end] == target) {
            return true;
        }
        return false;
    }
};