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