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。
1 package Algorithms.array; 2 3 public class MaxProduct { 4 public static int maxProduct(int[] A) { 5 if (A == null || A.length == 0) { 6 return 0; 7 } 8 9 // record the max value in the last node. 10 int DMax = A[0]; 11 12 // record the min value in the last node. 13 int DMin = A[0]; 14 15 // This is very important, should recode the A[0] as the initial value. 16 int max = A[0]; 17 18 for (int i = 1; i < A.length; i++) { 19 int n1 = DMax * A[i]; 20 int n2 = DMin * A[i]; 21 22 // we can select the former nodes, or just discade them. 23 DMax = Math.max(A[i], Math.max(n1, n2)); 24 max = Math.max(max, DMax); 25 26 // we can select the former nodes, or just discade them. 27 DMin = Math.min(A[i], Math.min(n1, n2)); 28 } 29 30 return max; 31 } 32 33 /* 34 * 作法是找到连续的正数,不断相乘即可。 35 * */ 36 public static void main(String[] strs) { 37 int[] A = {2, 3, -2, 4}; 38 39 System.out.println(maxProduct(A)); 40 } 41 }
时间: 2024-10-27 19:59:32