作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 双重循环 动态规划 参考资料 日期 题目地址:https://leetcode.com/problems/maximum-product-subarray/description/ 题目描述 Given an integer array nums, find the contiguous subarray within an array (conta…

题目链接:Maximum Product Subarray solutions同步在github 题目很简单,给一个数组,求一个连续的子数组,使得数组元素之积最大.这是求连续最大子序列和的加强版,我们可以先看看求连续最大子序列和的题目maximum-subarray,这题不难,我们举个例子. 假设数组[1, 2, -4, 5, -1, 10],前两个相加后得到3,更新最大值(为3),然后再加上-4后,和变成-1了,这时我们发现如果-1去加上5,不如舍弃前面相加的sum,5单独重新开始继续往后相加…

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. 解题思路: 计算连续的积最大,由于会有负数出现,因此需要用两个int表示包含num…

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. 题目标签:Array, Dynamic Programming 题目给了我们一个nu…

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…

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. 找出最大的相乘的数字,很简单,代码还可以优化的地方很多.但是速度还可以. publi…

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. 分析:这个题目就是让你求连续的子数组的乘积的最大值. (1)在数组中没有0的情况下,…

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]…

原题地址 简单动态规划,跟最大子串和类似. 一维状态空间可以经过压缩变成常数空间. 代码: int maxProduct(int A[], int n) { ) ; ]; ]; ]; ; i >= ; i--) { int tmp = minp; minp = min(A[i], min(A[i] * minp, A[i] * maxp)); maxp = max(A[i], max(A[i] * tmp, A[i] * maxp)); res = max(res, maxp); } retur…