[LeetCode]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.

题目翻译：在不重复的循环有序数组（如0 1 2 3 4 5 6 7->4 5 6 7 0 1 2）中查找目标，返回下标，如果不存在则返回-1.

知识点：二分查找

难点：判断目标在左边[target,mid-1]还是[mid+1,end]-->用IsPossible（int start,int end,int target)判断。

AC代码：

class Solution {public:    bool IsPossible(int start,int end,int target)    {        if(start==target||end==target)            return true;        if(start<end)        {            return(target>start&&target<end);        }else if(start>end)        {            return(target>start||target<end);        }else{            return false;        }    }    int search(int A[], int n, int target) {    int left=0,right=n-1;    while(left<=right)    {        int mid=(left+right)/2;        if(A[mid]==target)            return mid;        if(IsPossible(A[0],A[mid],target))        {            right=mid-1;/*target在mid左边*/        }        else        {            left=mid+1;/*target在mid右边*/        }    }    return -1;    }};