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;i<nums.length;i++)
{
int 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 <iostream>
#include <algorithm>
#include <vector>

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;
}
};