题目：
Search in Rotated Sorted Array
Total Accepted: 81090 Total Submissions: 277272 Difficulty: Hard
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.
思路：
数组被翻转，我们以中间元素为核心，把它分为两种情况，一种情况是中间元素的左边是有序的，一种是中间元素的右边是有序的。比如{ 0 1 2 4 5 6 7 }被翻转为{ 6 7 0 1 2 4 5 }或{ 2 4 5 6 7 0 1 }。
所以每次判断我们都取轻避重，总是跟有序的这边比较。
packagearray;publicclassSearchInRotatedSortedArray {publicintsearch(int[] nums,inttarget) {intlen = 0;if(nums ==null|| (len = nums.length) == 0)return0;//{ 0 1 2 4 5 6 7 }//1. { 6 7 0 1 2 4 5 }//2. { 2 4 5 6 7 0 1 }intl = 0;intr = len - 1;while(l <=r) {intmid = (l + r) / 2;if(nums[mid] == target)returnmid;if(nums[mid] <nums[r]) {if(target > nums[mid] && target <=nums[r])
l= mid + 1;elser= mid - 1;
}else{if(nums[l] <= target && target <nums[mid])
r= mid - 1;elsel= mid + 1;
}
}return-1;
}publicstaticvoidmain(String[] args) {//TODO Auto-generated method stubint[] nums = { 4, 5, 6, 7, 0, 1, 2};
SearchInRotatedSortedArray s=newSearchInRotatedSortedArray();
System.out.PRintln(s.search(nums,2));
}
}