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MaxProduct.java
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66 lines (50 loc) · 1.69 KB
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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;
}
/*
* 2014.12.20 Redo
* */
public int maxProduct2(int[] A) {
if (A == null || A.length == 0) {
return 0;
}
int max = 1;
int min = 1;
int ret = Integer.MIN_VALUE;
for (int i = 0; i < A.length; i++) {
int n1 = max * A[i];
int n2 = min * A[i];
max = Math.max(A[i], Math.max(n1, n2));
min = Math.min(A[i], Math.min(n1, n2));
ret = Math.max(max, ret);
}
return ret;
}
/*
* 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));
}
}