33. Search in Rotated Sorted Array

题目：

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.

链接： http://leetcode.com/problems/search-in-rotated-sorted-array/

题解：Rotated array中查找值，使用Binary Search，要注意数组是向左shift或者是向右shift，保持在一个单调递增的部分进行查找。

简单的三种情况 - 1) 不平移   1234567

　　　　　　　　2) 向左3位  4567123

　　　　　　     3) 向右3位   5671234 

时间复杂度 O(logn)， 空间复杂度 O(1)

public class Solution {

    public int search(int[] A, int target) {

        if(A == null || A.length == 0)

            return -1;

        int left = 0, right = A.length - 1;

      

        while(left <= right){

            int mid = right - (right - left) / 2;

            if(A[mid] == target)

                return mid;

            else if(A[mid] < A[left]){

               if(target > A[mid] && target <= A[right])

                   left = mid + 1;

               else

                   right = mid - 1;

            } else  {

               if(target < A[mid] && target >= A[left])

                   right = mid - 1;

               else

                   left = mid + 1;

            }

        }

        

        return -1;

    }

}

 

测试：