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MaxProduct.java

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package Algorithms.array;

public class MaxProduct {

public static int maxProduct(int[] A) {

if (A == null || A.length == 0) {

return 0;

}

// record the max value in the last node.

int DMax = A[0];

// record the min value in the last node.

int DMin = A[0];

// This is very important, should recode the A[0] as the initial value.

int max = A[0];

for (int i = 1; i < A.length; i++) {

int n1 = DMax * A[i];

int n2 = DMin * A[i];

// we can select the former nodes, or just discade them.

DMax = Math.max(A[i], Math.max(n1, n2));

max = Math.max(max, DMax);

// we can select the former nodes, or just discade them.

DMin = Math.min(A[i], Math.min(n1, n2));

}

return max;

}

/*

* 作法是找到连续的正数，不断相乘即可。

* */

public static void main(String[] strs) {

int[] A = {2, 3, -2, 4};

System.out.println(maxProduct(A));

}

}