LeetCode: Maximum Product Subarray 解题报告

Maximum Product Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest product.
For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.
Array Dynamic Programming

 

SOLUTION 1

使用DP来做：

因为有正负值好几种情况。所以我们计算当前节点最大值，最小值时，应该考虑前一位置的最大值，最大值几种情况。（例如：如果当前为-2， 前一个位置最小值为-6，最大值为8，那么当前最大值应该是-2*-6 = 12）

对于以index位置结尾的连续子串来说，计算最大，最小值可以三种选择：

1. 当前值* 前一位置的最大值。

2. 当前值* 前一位置的最小值。

3. 丢弃前一伴置的所有的值

我们对这三项取最大值，最小值，就可以得到当前的最大乘积，最小乘积。

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));
    }
}

2014.12.20  Redo

public class Solution {
    public int maxProduct(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;
    }
}

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/MaxProduct.java