leetcode 152. Maximum Product Subarray 最大连乘子序列 + 很棒的动态规划DP做法

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.

题意很简单，这里不就不说了，建议和leetcode 123. Best Time to Buy and Sell Stock III 一起学习。

建议和leetcode 560. Subarray Sum Equals K 动态规划DP子数组求和 和 leetcode 713. Subarray Product Less Than K 移动窗口 一起学习

就是使用DP动态求解，我也没想到是这么做得。

代码如下：

/*
 * 记录当前最大, 最小值. 因为遇到负数时, 与最小值的product可能成为最大值.
 * 
 * */
public class Solution 
{
    public int maxProduct(int[] nums) 
    {
        if(nums==null || nums.length<=0)
            return 0;

        int maxPro=nums[0] , minPro=nums[0] , maxRes=nums[0];
        for(int i=1;iint a = maxPro*nums[i];
            int b = minPro*nums[i];
            maxPro = Math.max(Math.max(a, b), nums[i]);
            minPro = Math.min(Math.min(a, b), nums[i]);
            maxRes = Math.max(maxRes, maxPro);
        }
        return maxRes;
    }
}

下面是C++的做法，就是做一个动态规划

代码如下：

#include 
#include 
#include 

using namespace std;

class Solution 
{
public:
    int maxProduct(vector<int>& nums) 
    {
        if (nums.size() <= 0)
            return 0;

        int maxPro = nums[0], minPro = nums[0], maxRes = nums[0];
        for (int i = 1; i < nums.size(); i++)
        {
            int a = maxPro*nums[i];
            int b = minPro*nums[i];
            maxPro = max(max(a, b),nums[i]);
            minPro = min(min(a, b), nums[i]);
            maxRes = max(maxRes, maxPro);
        }
        return maxRes;
    }
};