Maximum Product Subarray

https://oj.leetcode.com/problems/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.

解题思路：

这道题一看就是比较典型的动态规划题目。但和前面的max sum subarray相比，更为巧妙。

令子问题dp[i]为A[i]结尾的数组的最大乘积，状态转移方程就是dp[i] = max(A[i] * dp[i - 1], A[i])。不对，这里发现如果A[i]为负数的话，那么dp[i]就不行了，还必须记录一个A[i-1]结尾的数组的最小乘积，于是我们有第二个dp_min[i]。这样从低至上计算下来。

当然，这里完全可以不考虑A[i]的正负情况，直接dp_max[i] = max(A[i] * dp_max[i - 1], A[i], A[i] * dp_min[i - 1])，dp_min的计算同理。代码更为简洁，时间复杂度不变。

public class Solution {

    public int maxProduct(int[] A) {

        int[] dp_max = new int[A.length];

        int[] dp_min = new int[A.length];

        dp_max[0] = A[0];

        dp_min[0] = A[0];

        

        int max = A[0];

        int min = A[0];

        

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

            if(A[i] > 0){

                dp_max[i] = Math.max(A[i] * dp_max[i - 1], A[i]);

                dp_min[i] = Math.min(A[i] * dp_min[i - 1], A[i]);

            }

            if(A[i] < 0){

                dp_max[i] = Math.max(A[i] * dp_min[i - 1], A[i]);

                dp_min[i] = Math.min(A[i] * dp_max[i - 1], A[i]);

            }

            if(A[i] == 0){

                dp_max[i] = 0;

                dp_min[i] = 0;

            }

            if(dp_max[i] > max){

                max = dp_max[i];

            }

            if(dp_min[i] < min){

                min = dp_min[i];

            }

        }

        return max;

    }

}