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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：
33. Search in Rotated Sorted Array
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原文地址：http://www.cnblogs.com/yrbbest/p/4435871.html
