63. Search in Rotated Sorted Array II [LintCode]

Follow up forSearch in Rotated Sorted Array:

What ifduplicatesare 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, returntrue.
Given[1, 1, 1, 1, 1, 1]and target =0, returnfalse.

类似于Find Minimum in Rotated Sorted Array II，遇到A[start] == A[mid]的情况下不知道是2, 2, 2, 0, 1, 2 还是 2, 3, 4, 2, 2, 2的情况，无法判断指针移动情况，因此要指针一格一格移动来将重复元素过滤到单边。

public class Solution {
    /*
     * @param A: an integer ratated sorted array and duplicates are allowed
     * @param target: An integer
     * @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[start] < A[mid]) {
                if (A[start] <= target && target <= A[mid]) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else if (A[start] > A[mid]) {
                if (A[mid] <= target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            } else {
                start++;
            }
        }
        if (A[start] == target || A[end] == target) {
            return true;
        }
        return false;
    }
}