LeetCode OJ--Search in Rotated Sorted Array II

http://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/

如果在数组中有重复的元素，则不一定说必定前面或者后面的一半是有序的，所以有个while循环的处理。

#include <iostream>
using namespace std;
  
class Solution {
public:
    bool binarySearch(int A[],int target,int begin,int end)
    {
        int mid = (begin+end)/2;

        if(target == A[mid])
            return true;
        if(target == A[begin])
            return true;
        if(target == A[end])
            return true;
        if(begin>=end)
            return false;

        while(A[begin] == A[mid])
            begin++;
        if(A[begin]<A[mid])//前面有序
        {
            if(target>A[begin] && target<A[mid]) //在前面
                return binarySearch(A,target,begin,mid-1);
            else
                return binarySearch(A,target,mid+1,end); //在后面
        }
        else if(A[mid]<=A[end])//后面有序
        {
            while(A[mid] == A[end])
                end--;
            if(target>A[mid] && target<A[end]) //在后面
                return binarySearch(A,target,mid+1,end);
            else
                return binarySearch(A,target,begin,mid-1);//在前面
        }
        return false;
    }
    bool search(int A[], int n, int target) {
       return  binarySearch(A,target,0,n-1);
    }
};

int main() 
{
    Solution myS;

    int A[] = {1,3,1,1,1};
    bool ans = myS.search(A,5,3);
    cout<<ans<<endl;
    return 0;

}