Given an integern, break it into the sum ofkpositive integers, wherek >= 2, and maximize the product of those integers. Returnthe maximum product you can get. Example 1: Input:n = 2Output:1Explanation:2 = 1 + 1, 1 × 1 = 1. Example 2: Input:n = 10Output:36Explanation:10 = ...
343. Integer Break https://leetcode.com/problems/integer-break/description/ 解题思路: 双层循环 dp[i] = Math.max(dp[i], Math.max(dp[j], j) * Math.max(dp[i - j], i - j)); class Solution { public int integerBreak(int n) { } 转载于:ht......
LeetCode——Integer Break Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 ...
LeetCode题解 343.Integer Break 题目:Given a positive integern, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. For example, givenn= 2, return 1 (2 = 1 + 1); givenn= 10, return 36 (10 =...
class Solution: def integerBreak(self, n: int) -> int: #前 5 个数的最大乘积 tmp = [1, 2, 4, 6, 9] for i in range(5, n - 1): # 动态规划:第 i 个数 的最大乘积为 3 * 往前数第三个数 tmp.append(3 * tmp[i - 3]) ...
leetcode——integer break(343) 技术标签: 数据结构与算法 leetcode # -*- encoding: utf-8 -*- """ Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. Example 1: Input...
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leetcode 343. Integer Break 题目要求 Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36...
详见:https://leetcode.com/problems/integer-break/description/ C++: classSolution{public:intintegerBreak(intn){if(n==2||n==3){returnn-1;}intres=1;while(n>4){res*=3;n-=3;}returnres*n;}}; 1. 2. 3. 4. 5. 6. 7. 8. ...
classSolution{public:intmyAtoi(strings){if(s.empty()){return0;}inti=0,len=s.length();intres=0,firstChar=1;// // firstChar 表示正负号,默认为 '+',即 1intupperBound=INT_MAX/10;while(s[i]==' '){// 省去所有空格if(++i>=len){return0;}}if(s[i]=='-'){// 判断正负号firstCh...