Search in Rotated Sorted Array（二分查找）

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.

ps：这题和剑指offter的题目类似。一次AC

思路，由于没有重复数字，

1.我们先找出最大值在哪个位置。

2.然后拿target和A[0]做比较，若target>A[0],则在0~max_loc之间来二分查找数组；若target<A[0]，则在max_loc+1~n-1来查找数组。

代码：

class Solution {
public:
    int binarySearch(int start,int end,int A[],int target){
        int l=start;int r=end;int mid=;
        while(l<=r){
            mid=(l+r)/;
            if(A[mid]>target) r=mid-;
            else if(A[mid]<target) l=mid+;
            else return mid;
        }
        return -;
    }
    int findMaxValueLoction(int A[],int n){
        if(A[n-]>A[]) return n-;
        int l=;int r=n-;int mid=;
        while(l<=r){
            mid=(l+r)/;
            if(mid==) return mid;
            else if(mid!=&&A[mid]>A[]&&A[mid]>A[mid-]&&A[mid]<A[mid+]){
                l=mid+;
            }
            else if(mid!=&&A[mid]<A[]&&A[mid]>A[mid-]&&A[mid]<A[mid+]){
                r=mid-;
            }
            else if(mid!=&&A[mid]>A[mid-]&&A[mid]>A[mid+]){
                return mid;
            }else{
                return mid-;
            }
        }
        return -;
    }
    int search(int A[], int n, int target) {
        if(A==NULL||n==) return -;
        int max_loc=findMaxValueLoction(A,n);
        if(target>A[])
            return binarySearch(,max_loc,A,target);
        else if(target<A[])
            return binarySearch(max_loc+,n-,A,target);
        else
            return ;
    }
};