Search in Rotated Sorted Array

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.

分析：

显然用二分查找法喽

但是，当mid处于前半部分还是后半部分时对left和right的处理时不同，所以要分开处理

当mid处于前半部分时，如果left元素小于target，target也小于right元素，说明target在left和mid之间，则right = mid - 1；否则，left = mid + 1；

当mid处于后半部分时，如果right元素大于target,target也大于mid元素，说明target在mid和right之间，则left = mid + 1,否则，right = mid - 1

参考代码：

http://www.programcreek.com/2014/06/leetcode-search-in-rotated-sorted-array-java/

public class Solution {    public int search(int[] nums, int target) {    int left = 0;    int right= nums.length-1;     while(left<=right){        int mid = left + (right-left)/2;        if(target==nums[mid])            return mid;         if(nums[left]<=nums[mid]){            if(nums[left]<=target&& target<nums[mid]){                right=mid-1;            }else{                left=mid+1;            }        }else{            if(nums[mid]<target&& target<=nums[right]){                left=mid+1;            }else{                right=mid-1;            }        }        }    return -1;  }}