题目 Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. 给定正整数 n,找到若干个完全平方数(比如 1, 4, 9, 16, ...)使得它们的和等于 n.你需要让组成和的完全平方数的个数最少. Example 1: Input: n = 12 Output: 3 Explanation: 12 =…

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. Example 1: Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4. Example 2: Input: n = 13 Output: 2 Explanation: 13 = 4 + 9. Cr…

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. ExampleGiven n = 12, return 3 because 12 = 4 + 4 + 4Given n = 13, return 2 because 13 = 4 + 9 LeetCode上的原题,请参见我之前的博客Perfect Sq…

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. Credits:Special thanks…

https://leetcode.com/problems/perfect-squares/ Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, retur…

一原题 Given a positive integer n, find the least number of perfect square numbers (, , , , ...) which sum to n. For example, given n = , because = + + ; given n = , because = + . 二.中文讲解 .这道题说是给我们一个正整数,求它最少能由几个完全平方数组成.这道题是考察四平方和定理 .根据四平方和定理,任意一个正整数均可表示为…

给定正整数 n,找到若干个完全平方数(比如 1, 4, 9, 16, ...)使得它们的和等于 n.你需要让组成和的完全平方数的个数最少. 示例 1: 输入: n = 12 输出: 3 解释: 12 = 4 + 4 + 4. 示例 2: 输入: n = 13 输出: 2 解释: 13 = 4 + 9. class Solution { public: int numSquares(int n) { vector<int> squares; for(int i = 1; i * i <=…

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16 ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. 解析: 利用动态规划解决此问题:对于要求的当前节…

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. Example 1: Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4. Example 2: Input: n = 13 Output: 2 Explanation: 13 = 4 + 9.---…

原题链接 题意: 给一个非整数,算出其最少可以由几个完全平方数组成(1,4,9,16……) 思路: 可以得到一个状态转移方程  dp[i] = min(dp[i], dp[i - j * j] + ); 代码如下: Runtime: 60 ms, faster than 69.83% 有点慢 class Solution { public: int numSquares(int n) { ]; ; i <= n; i++) { dp[i] = i; ; j * j <= i; j++) { d…