Maximum Product Subarray

The maximum product of sub-arrays in [1,n] can be divided by 3 cases:

  1. A[n] is the maximum product of all sub-arrays in [1, n].
  2. The array which has the maximum product is end by A[n].
  3. The array of maximum product is not including A[n]. 

Thus the result can be expressed as

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result = max(case1, case2, case3)

The second situation is not normal. If A[n] is positive, it can be got by prePositiveMax[n-1], if A[n] is negative, it can be got by preNegativeMin[n-1].

 

So one brute method is: 

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prePositiveMax[n] = max(prePositiveMax[n-1]*A[n], preNegativeMin[n-1]*A[n], A[n]);
  
preNegativeMin[n] = min(prePositiveMax[n-1]*A[n], preNegativeMin[n-1]*A[n], A[n]);

 

A more accurate method is:  

We define pEnd[i]: the maximum non-negative product of subarray with A[i]

We define nEnd[i]: the minimum non-positive product of subarray with A[i]

 

In fact, here we use pEnd, nEnd intead of prePositiveMax, preNegativeMin, thus we have  

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if(A[i] > 0){    
    pEnd[i] = max(A[i], pEnd[i-1]*A[i]);    
    nEnd[i] = nEnd[i] * A[i];
}else{    
    pEnd[i] = nEnd[i] * A[i];    
    nEnd[i] = min(A[i], pEnd[i]*A[i]);
}

 

Then, we can simplify as  

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if(A[i] < 0) swap(pEnd, nEnd);
 
pEnd = max(pEnd*A[i], A[i]);
nEnd = min(nEnd*A[i], A[i]);

 

So we conclude the whole code 

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int maxProduct(int A[], int n) {
    if(1 == n) return A[0];
     
    int pEnd, nEnd, res;
     
    pEnd = nEnd = res = 0;
     
    for(int i = 0; i < n; ++i){
        if(A[i] < 0) swap(&pEnd, &nEnd);
         
        pEnd = max(pEnd*A[i], A[i]);
        nEnd = min(nEnd*A[i], A[i]);
         
        if(res < pEnd) res = pEnd; //from res = max(res, pEnd)
    }
     
    return res;
}

 


 

The complete code is:

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class Solution {
private:
    int max(int a, int b){
        return a > b ? a : b;
    }
    int min(int a, int b){
        return a > b ? b : a;
    }
    void swap(int* a, int* b){
        *a = *a + *b;
        *b = *a - *b;
        *a = *a - *b;
    }
     
public:
    int maxProduct(int A[], int n) {
        if(1 == n) return A[0];
         
        int pEnd, nEnd, res;
         
        pEnd = nEnd = res = 0;
         
        for(int i = 0; i < n; ++i){
            if(A[i] < 0) swap(&pEnd, &nEnd);
             
            pEnd = max(pEnd*A[i], A[i]);
            nEnd = min(nEnd*A[i], A[i]);
             
            if(res < pEnd) res = pEnd;
        }
         
        return res;
         
    }
};

 

posted @   kid551  阅读(299)  评论(0编辑  收藏  举报
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