62. Search in Rotated Sorted Array [LintCode]

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e.,0 1 2 4 5 6 7might become4 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]andtarget=1, return2.

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

用A[start] 和 A[mid]的关系判断哪一段是递增区间，判断出哪一段是递增区间之后，可以判断target是否在递增区间内，从而调整start或者end指针达到减半目的。

public class Solution {
    /*
     * @param A: an integer rotated sorted array
     * @param target: an integer to be searched
     * @return: an integer
     */
    public int search(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0) {
            return -1;
        }
        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[mid] <= target && target <= A[end]) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
        }
        if (A[start] == target) {
            return start;
        }
        if (A[end] == target) {
            return end;
        }
        return -1;
    }
}