leetcode 题解：Search in Rotated Sorted Array II （旋转已排序数组查找2）

题目：

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

说明：

     1）和1比只是有重复的数字，整体仍采用二分查找

     2）方法二 ：

实现：

   一、我的实现：

class Solution {
public:
    bool search(int A[], int n, int target) {
        if(n==0||n==1&&A[0]!=target) return false;
        if(A[0]==target) return true;
        int i=1;
        while(A[i-1]<=A[i])  i++;
        
        bool pre=binary_search(A,0,i,target);
        bool pos=binary_search(A,i,n-i,target);
        return pre==false?pos:pre;
    }
private:
    bool binary_search(int *B,int lo,int len,int goal)
    {
        int low=lo;
        int high=lo+len-1;
        while(low<=high)
        {
            int middle=(low+high)/2;
            if(goal==B[middle])//找到，返回index
               return true;
            else if(B[middle]<goal)//在右边
               low=middle+1;
            else//在左边
               high=middle-1;
        }
        return false;//没有，返回-1
    }
};

二、网上开源实现：

class Solution {
public:
    bool search(int A[], int n, int target) {
         int low=0;
         int high=n-1;
         while(low<=high)
         {
            const int middle=(low+high)/2;
            if(A[middle]==target) return true;
            if(A[low]<A[middle])
            {
               if(A[low]<=target&&target<A[middle])
                  high=middle-1;
               else
                  low=middle+1;
            }
            else if(A[low]>A[middle])
            {
               if(A[middle]<target&&target<=A[high])
                  low=middle+1;
               else 
                  high=middle-1;
            }
            else
                low++;
         }
         return false;
    }
};