DP Leetcode - Maximum Product Subarray

最近一直忙着写paper，很久没做题，一下子把题目搞复杂了。。思路理清楚了非常简单，每次只需更新2个值：当前子序列最大乘积和当前子序列的最小乘积。最大乘积被更新有三种可能：当前A[i]>0,乘以前面最大的数（>0),得到新的最大乘积；当前A[i]<0,乘以前面最小的数（<0),得到新的最大乘积；A[i]它自己>0,（A[i-1]==0。最小乘积同理。。

class Solution {public:int Max(int a, int b, int c){int max = -1 << 30;if (a >= b)max = a;elsemax = b;if (c > max)max = c;return max;}int Min(int a, int b, int c){int min = 1 << 30;if (a <= b)min = a;elsemin = b;if (c < min)min = c;return min;}int maxProduct(int A[], int n) {int maxPPrev=A[0], maxP;int minPPrev = A[0], minP;int max = A[0];for (int i = 1; i < n; i++){maxP = Max(maxPPrev * A[i], minPPrev * A[i], A[i]);minP = Min(maxPPrev * A[i], minPPrev * A[i], A[i]);maxPPrev = maxP;minPPrev = minP;if (maxPPrev > max)max = maxPPrev;if (minPPrev > max)max = minPPrev;}return max;}};