Problem

Suppose a sorted array 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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Example
For [4, 5, 1, 2, 3] and target=1, return 2.

For [4, 5, 1, 2, 3] and target=0, return -1.

思路

• 因为duplicate，导致了A[mid] 可能等于A[end], 从而影响判断

• 解决思路：创造条件使之变成 I , 也就是跳过相等的元素。 判断加上 A[mid] == A[end] 情况的判断, end--, 直到找到不相同才进行二分

    public boolean search(int[] A, int target) {
        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;
            }

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