leetcode -- Maximum Product Subarray -- 重点
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https://leetcode.com/problems/maximum-product-subarray/
参考思路:http://yucoding.blogspot.hk/2014/10/leetcode-quesion-maximum-product.html
其实这里与Maximum subarray思路很像,都是dp经典。对max[i]的定义这里与Maximum subarray一致,只是要再定义一个min[i].
这里我们其实只是讨论max[i]与max[i-1]以及min[i-1]的关系,max[i]必然处于case 1,2两种情况之中
case 1: 首先我们考虑以A[i]为结尾的subarray只有A[i]的情况。
就是求max[i]时不能利用max[i-1], 即A[i-1] == 0, 导致max[i-1] == min[i-1] == 0的情况。这个时候,max[i] = A[i],也有可能是A[i] >0, max[i-1] < 0, 那么也是A[i]本身。
case 2: 再考虑以A[i]为结尾的subarray,不只有A[i]的情况
当A[i] > 0, max[i-1] > 0时,max[i] = A[i] * max[i-1]。
当A[i] > 0, max[i-1] < 0时,max[i] = A[i]。(这个其实是属于case1,只有A[i]的情况)
当A[i]<0时,为了求max[i]
- min[i-1]>0, max[i-1]>0: max[i] = A[i]*min[i-1]
- min[i-1]<0, max[i-1]<0: max[i] = A[i]*min[i-1]
- min[i-1]<0, max[i-1]>0: max[i] = A[i]*min[i-1]
其实也不难理解负数乘以一个较小数要大于乘以一个较大数。例如 given
所以max[i] = A[i]*min[i-1] when A[i] <0
最后max[i] = max(A[i]*max[i-1], A[i]*min[i-1], A[i]). 对于min[i], 对上式取min即可,同样的道理推导
最终取得max[i]中最大的元素即可
class Solution(object): def maxProduct(self, nums): """ :type nums: List[int] :rtype: int """ if len(nums) < 2: return None res = nums[0] minp = nums[0] maxp = nums[0] for a in nums[1:]: tmp_maxp = maxp tmp_minp = minp maxp = max(max(tmp_maxp*a, tmp_minp*a), a) minp = min(min(tmp_maxp*a, tmp_minp*a), a) print (maxp, minp) res = max(res, maxp) return res
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