Maximum Product Subarray

最新推荐文章于 2021-07-19 16:22:36 发布

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最新推荐文章于 2021-07-19 16:22:36 发布

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原文链接：http://www.cnblogs.com/kid551/p/4113020.html

本文介绍了一种求解最大子数组乘积的算法，包括基本情况的处理、两种方法的对比，以及代码实现细节。

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

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

Thus the result can be expressed as

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:

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

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

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

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:

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;
        
    }
};

转载于:https://www.cnblogs.com/kid551/p/4113020.html