标签：style blog ar color sp for on div log
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.
找数字连续最大乘积子序列。
思路：这个麻烦在有负数和0，我的方法，如果有0，一切都设为初始值。
对于两个0之间的数若有奇数个负数，那则有两种情况，第一种是不要第一个负数和之前的值，第二种是不要最后一个负数和之后的值，用negtiveFront和negtiveBack表示。没有负数就是不要第一个负数和之前的值的情况。
int maxProduct(int A[], int n) { if(n == 0) return 0; int MaxAns = A[0]; int negtiveFront = (A[0] == 0) ? 1 : A[0]; int negtiveBack = (A[0] < 0) ? 1 : 0; for(int i = 1; i < n; i++) { if(A[i] == 0) { MaxAns = (MaxAns > 0) ? MaxAns : 0; negtiveFront = 1; negtiveBack = 0; } else if(A[i] < 0) { negtiveFront *= A[i]; MaxAns = max(negtiveFront, MaxAns); if(negtiveBack == 0) { negtiveBack = 1; } else { negtiveBack *= A[i]; MaxAns = max(negtiveBack, MaxAns); } } else { negtiveFront *= A[i]; negtiveBack *= A[i]; MaxAns = max(negtiveFront, MaxAns); if(negtiveBack > 0) { MaxAns = max(negtiveBack, MaxAns); } } } return MaxAns; }
答案的思路：同时维护包括当前数字A[k]的最大值f(k)和最小值g(k)
f(k) = max( f(k-1) * A[k], A[k], g(k-1) * A[k] ) g(k) = min( g(k-1) * A[k], A[k], f(k-1) * A[k] )
再用一个变量Ans存储所有f(k)中最大的数字就可以了
int maxProduct2(int A[], int n) { if(n == 0) return 0; int MaxAns = A[0]; //包括当前A【i】的连续最大乘积 int MinAns = A[0]; //包括当前A【i】的连续最小乘积 int MaxSoFar = A[0]; //整个数组的最大乘积 for(int i = 1; i < n; i++) { int MaxAnsTmp = MaxAns; int MinAnsTmp = MinAns; MaxAns = max(MaxAnsTmp * A[i], max(MinAnsTmp * A[i], A[i])); MinAns = min(MinAnsTmp * A[i], min(MaxAnsTmp * A[i], A[i])); MaxSoFar = max(MaxSoFar, MaxAns); } return MaxSoFar; }
【leetcode】 Unique Binary Search Trees (middle)☆
标签：style blog ar color sp for on div log
原文地址：http://www.cnblogs.com/dplearning/p/4158432.html
