[LeetCode] Search in Rotated Sorted Array II 二分搜索

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.

 

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  Array Binary Search

 

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　　这个是在 前一道的基础上改过来的，主要是 left middle right 相同的时候，递归为左右部分的并，其他部分不变：

class Solution {

public:

    bool search(int A[], int n, int target) {

        if(n==1)    return A[0]==target?true:false;

        return help(A,0,n-1,target);

    }

    int help(int* & A,int l,int r,int &target)

    {

        if(target<A[l]&&target>A[r])    return false;

        if(target==A[l])    return true;

        if(target==A[r])    return true;

        if(r-l==1)  return false;

        int m = (l+r) /2;

        if(target==A[m])    return true;

        if(A[l]<A[m]){

            if(A[l]<target&&target<A[m])    return help(A,l,m,target);

            return help(A,m,r,target);

        }

        else if(A[l]>A[m]){

            if(A[m]<target&&target<A[r])    return help(A,m,r,target);

            return help(A,l,m,target);

        }

        else{

            return help(A,l,m,target)||help(A,m,r,target);

        }

    }

};