Suppose an array sorted in ascending order 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.
假设有一个排序的按未知的旋转轴旋转的数组(比如，0 1 2 4 5 6 7 可能成为4 5 6 7 0 1 2)。给定一个目标值进行搜索，如果在数组中找到目标值返回数组中的索引位置，否则返回-1。
1 public int search(int[] nums, int target) { 2 int n = nums.length; 3 int lo=0,hi=n-1; 4 // find the index of the smallest value using binary search. 5 // Loop will terminate since mid < hi, and lo or hi will shrink by at least 1. 6 // Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated. 7 while(lo<hi){ 8 int mid=(lo+hi)/2; 9 if(nums[mid]>nums[hi]) lo=mid+1; 10 else hi=mid; 11 } 12 // lo==hi is the index of the smallest value and also the number of places rotated. 13 int rot=lo; 14 lo=0;hi=n-1; 15 // The usual binary search and accounting for rotation. 16 while(lo<=hi){ 17 int mid=(lo+hi)/2; 18 int realmid=(mid+rot)%n; 19 if(nums[realmid]==target)return realmid; 20 if(nums[realmid]<target)lo=mid+1; 21 else hi=mid-1; 22 } 23 return -1; 24 }
33. Search in Rotated Sorted Array
原文：http://www.cnblogs.com/wzj4858/p/7675385.html