Maximum Product Subarray JAVA实现

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.

package com.edu.leetcode;

import javax.xml.transform.Templates;

public class MaximumProductSubarray {

    public int maxProducts(int[] A) {
        if(A.length==1){                                      //如果数组中只有一个元素直接输出
            return A[0];
        }
        int[] oneValues = new int[A.length];            //辅助空间1，记录从头或者从0开始到当前位置的乘积
        int[] twoValues=new int[A.length];            //辅助空间2，从头或者从0开始遇到第一负数时，后面的值重新从当前位置开始
        oneValues[0]=A[0];
        twoValues[0]=A[0];
        boolean first =true;                                    //从头或者从0开始，判断是否遇到了负数
        for(int i=1;i<A.length;i++){
            if(oneValues[i-1]==0){
                oneValues[i]=A[i];
                twoValues[i]=A[i];
                first=true;
            }
            else{
                if(A[i-1]<=-1&&first){                 //当前面的一个数是从头或者从0开始的第一个负数时，将twoValues[i]的值赋值为A[i]
                    twoValues[i]=A[i];
                    oneValues[i]=oneValues[i-1]*A[i];
                    first=false;
                }
                else{                                 
                    oneValues[i]=oneValues[i-1]*A[i];
                    twoValues[i]=twoValues[i-1]*A[i];
                }
            }
        }
        int maxValue=oneValues[0];
        for(int i=0;i<oneValues.length;i++){
            if(oneValues[i]>maxValue)
                maxValue=oneValues[i];
            if(twoValues[i]>maxValue)
                maxValue=twoValues[i];
        }    
        return maxValue;
    } 

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        MaximumProductSubarray mps = new MaximumProductSubarray();
        int[] s = {2,3,-1};
        System.out.println(mps.maxProducts(s));
        
    }

}