lintcode62- Search in Rotated Sorted Array- medium

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.

Challenge 

O(logN) time

halfhalf二分法，每次和最后一个数以及target比一比，决定去左半还是右半。

和find the minimum类似，只是分类思想分的得更细一点。先考虑切到上面还是下面，再考虑target在上下产生的不同寻找方向。

public class Solution {
    public int search(int[] A, int target) {
        if (A == null || A.length == 0) {
            return -1;
        }

        int start = 0;
        int end = A.length - 1;
        int mid;
        
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return mid;
            }
            if (A[start] < A[mid]) {
                // situation 1, red line
                if (A[start] <= target && target <= A[mid]) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else {
                // situation 2, green line
                if (A[mid] <= target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
        } // while
        
        if (A[start] == target) {
            return start;
        }
        if (A[end] == target) {
            return end;
        }
        return -1;
    }
}