Search in Rotated Sorted Array II(lintcode 63)

Description

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.

Example

Given [1, 1, 0, 1, 1, 1] and target = 0, return true. Given [1, 1, 1, 1, 1, 1] and target = 0, return false.

Interface

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
    }
}

Solution

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0) {
            return false;
        }

        int start = 0;
        int end = A.length - 1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;

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

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

        return false;
    }
}