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=0;
        while(l<=r){
            mid=(l+r)/2;
            if(A[mid]>target) r=mid-1;
            else if(A[mid]<target) l=mid+1;
            else return mid;
        }
        return -1;
    }
    int findMaxValueLoction(int A[],int n){
        if(A[n-1]>A[0]) return n-1;
        int l=0;int r=n-1;int mid=0;
        while(l<=r){
            mid=(l+r)/2;
            if(mid==0) return mid;
            else if(mid!=0&&A[mid]>A[0]&&A[mid]>A[mid-1]&&A[mid]<A[mid+1]){
                l=mid+1;
            }
            else if(mid!=0&&A[mid]<A[0]&&A[mid]>A[mid-1]&&A[mid]<A[mid+1]){
                r=mid-1;
            }
            else if(mid!=0&&A[mid]>A[mid-1]&&A[mid]>A[mid+1]){
                return mid;
            }else{
                return mid-1;
            }
        }
        return -1;
    }
    int search(int A[], int n, int target) {
        if(A==NULL||n==0) return -1;
        int max_loc=findMaxValueLoction(A,n);
        if(target>A[0]) 
            return binarySearch(0,max_loc,A,target);
        else if(target<A[0]) 
            return binarySearch(max_loc+1,n-1,A,target);
        else 
            return 0;
    }
};