[LeetCode]Maximum Product Subarray
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今天准备再刷一遍leetcode,发现出了一道新题。刚开始想贪心一遍O(n)实现,没有想出来。于是从头开始想,暴力双循环,O(n*n)肯定没有问题。然后下午坐车的时候又想到了可以用二分法,后来终于又想出来了O(n)的实现,如下所示,不过怎么感觉自己的实现这么麻烦呢,对0考虑的太多了。后面贴的别人的算法就比较简洁。
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
.
int maxProduct(int A[], int n) { int res = A[0]; int nmax = A[0] < 0 ? A[0] : 0; int pmax = A[0] > 0 ? A[0] : 0; for (int i = 1; i < n; i++) { if (A[i] == 0) { nmax = pmax = 0; } else if (A[i] > 0) { nmax *= A[i]; pmax = pmax != 0 ? pmax * A[i] : A[i]; } else { swap(pmax, nmax); nmax = nmax != 0 ? nmax * A[i] : A[i]; pmax *= A[i]; } res = max(res, pmax); } return res;}
http://blog.csdn.net/sbitswc/article/details/39546719
public int maxProduct2(int[] A) {if(A.length<=0) return 0;if(A.length==1) return A[0];int curMax = A[0];int curMin = A[0];int ans = A[0];for(int i=1;i<A.length;i++){int temp = curMin *A[i];curMin = Math.min(A[i], Math.min(temp, curMax*A[i]));curMax = Math.max(A[i], Math.max(temp, curMax*A[i]));ans = Math.max(ans, curMax);}return ans;}
思路也很清晰:
Analysis:
similar like Maximum Subarray question
difference is the max value could be get from 3 situations
current maxValue * A[i] if A[i]>0
current minValue * A[i] if A[i]<0
A[i]
We need to record current maxValue, current minValue and update them every time get the new product
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