LeetCode Maximum Product Subarray Description Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of the maximum positive product involving consecutive terms of S. If you cannot find a positive sequence, you…

Maximum Product Subarray Title: 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. 对于Product…

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. 这个求最大子数组乘积问题是由最大子数组之和问题演变而来,但是却比求最大子数组之和要复…

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. 子数组乘积最大 class Solution { public: int maxPr…

Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product. Example 1: Input: [2,3,-2,4] Output: 6 Explanation: [2,3] has the largest product 6. Example 2: Input: [-2,0,-1]…

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. Hide Tags Array Dynamic Programming      …

原题地址:https://oj.leetcode.com/problems/maximum-product-subarray/ 解题思路:主要需要考虑负负得正这种情况,比如之前的最小值是一个负数,再乘以一个负数就有可能成为一个很大的正数. 代码: class Solution: # @param A, a list of integers # @return an integer def maxProduct(self, A): if len(A) == 0: return 0 min_tmp…

LeetCode 新题又更新了.求:最大子数组乘积. https://oj.leetcode.com/problems/maximum-product-subarray/ 题目分析:求一个数组,连续子数组的最大乘积. 解题思路:最简单的思路就是3重循环.求解productArray[i][j]的值(productArray[i][j]为A[i]到A[j]的乘积),然后记录当中最大值,算法时间复杂度O(n3).必定TLE. 第一次优化,动态规划.求解:productArray[i][j]的时候不用…

题意:给一个size大于0的序列,求最大的连续子序列之积.(有正数,负数,0) 思路:正确分析这三种数.0把不同的可能为答案的子序列给隔开了,所以其实可以以0为分隔线将他们拆成多个序列来进行求积,这样就没有0了. 接着是负数,负数如果遇到一个负数,可能反而比那个正数要大,所以正负数都要保存,遍历一次即可.在奇数个负数时,其实可能的只有2种:(1)包含最前面一个负数的序列(2)包含最后面一个负数的序列(当然不包含最前面1个,同理(1)也是). class Solution { public: in…

近期一直忙着写paper,非常久没做题,一下子把题目搞复杂了..思路理清楚了非常easy,每次仅仅需更新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,乘曾经面最大的数(>0),得到新的最大乘积:当前A[i]<0,乘曾经面最小的数(<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) { in…