Leetcode: Maximum Product Subarray

题目：
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.

这里的Product是乘积的意思。这道题的tag是Dynamic Programming，所以同样是动态规划的思想，找出局部和全局的递推关系。

思路：这道题和上一道题：Maximum Subarray思路差不多。Maximum Subarray是求子数组和的最大值，这道题是求子数组乘积的最大值。计算最大的乘积同样要考虑负数的情况：一个很小的负数乘以一个负数，可能是一个很大的正数。所以这道题我们要设置两个局部最优变量，一个保存局部最大值，一个保存局部最小值（其实只有当这个局部最小值是负数的时候，才真正起作用）。
它们有如下关系：
copyMax = localMax
localMax = max(max(localMax * A[i], localMin * A[i]), A[i])
localMin = min(min(copyMax * A[i], localMin * A[i]), A[i])
globalMax = max(localMax, globalMax)

C++代码：

class Solution
{
public:
    int maxProduct(int A[], int n)
    {
        if (n <= 0) return 0;
        int globalMax = A[0];
        int localMax = A[0];
        int localMin = A[0];
        int copyMax = localMax;
        for (int i = 1; i < n; i++)
        {
            copyMax = localMax;
            localMax = max(max(localMax * A[i], localMin * A[i]), A[i]);
            localMin = min(min(copyMax * A[i], localMin * A[i]), A[i]);
            globalMax = max(localMax, globalMax);
        }
        return globalMax;
    }
};

C#代码：

public class Solution
{
    public int MaxProduct(int[] A) 
    {
        if (A.Length <= 0) return 0;
        int globalMax = A[0];
        int localMax = A[0];
        int localMin = A[0];
        int copyMax = localMax;
        for (int i = 1; i < A.Length; i++)
        {
            copyMax = localMax;
            localMax = Math.Max(Math.Max(localMax * A[i], localMin * A[i]), A[i]);
            localMin = Math.Min(Math.Min(copyMax * A[i], localMin * A[i]), A[i]);
            globalMax = Math.Max(globalMax, localMax);
        }
        return globalMax;
    }
}

Python代码：

class Solution:
    # @param A, a list of integers
    # @return an integer
    def maxProduct(self, A):
        size = len(A)
        if size <= 0:
            return 0
        globalMax = A[0]
        localMax = A[0]
        localMin = A[0]
        copyMax = localMax
        for i in range(1, size):
            copyMax = localMax
            localMax = max(max(localMax * A[i], localMin * A[i]), A[i])
            localMin = min(min(copyMax * A[i], localMin * A[i]), A[i])
            globalMax = max(globalMax, localMax)
        return globalMax