leetcode@ [279]Perfect Squares
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, return 2 because 13 = 4 + 9.
class Solution {
public:
int getMaximumSquare(int n){
int ret = (int)sqrt(n);
return ret;
}
int dfs(int load, vector<int> &dp){
if(dp[load]) return dp[load];
if(load == ){
return ;
}
int next = getMaximumSquare(load);
int Min = numeric_limits<int>::max();
for(int v=next;v>=;v--){
if(load-v*v >= ){
Min = min(Min, dfs(load-v*v, dp));
}
}
dp[load] = Min + ;
return dp[load];
}
int numSquares(int n) {
vector<int> dp(n+);
for(int i=;i<dp.size();++i) dp[i] = ;
dp[n] = dfs(n, dp);
return dp[n];
}
};
leetcode@ [279]Perfect Squares的更多相关文章
- [LeetCode] 279. Perfect Squares 完全平方数
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 1 ...
- (BFS) leetcode 279. Perfect Squares
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 1 ...
- [leetcode] #279 Perfect Squares (medium)
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] + ) ...
- [LeetCode 279.] Perfect Squres
LeetCode 279. Perfect Squres DP 是笨办法中的高效办法,又是一道可以被好办法打败的 DP 题. 题目描述 Given a positive integer n, find ...
- [LeetCode] 0279. Perfect Squares 完全平方数
题目 Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9 ...
- 【LeetCode】279. Perfect Squares 解题报告(C++ & Java)
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 四平方和定理 动态规划 日期 题目地址:https: ...
- 【leetcode】Perfect Squares (#279)
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 1 ...
- 279. Perfect Squares
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 1 ...
- 279. Perfect Squares(动态规划)
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 1 ...
随机推荐
- setjump 和 longjump
goto语句可以用于同一个函数内异常处理,不幸的是,goto是本地的,它只能跳到所在函数内部的标号上.为了解决这个限制,C函数库提供了setjmp()和longjmp()函数,它们分别承担非局部标号和 ...
- Side-by-side assembly
Side-by-side technology is a standard for executable files in Windows 98 Second Edition, Windows 200 ...
- Android TextView结合SpannableString使用
super.onCreate(savedInstanceState); TextView txtInfo = new TextView(this); SpannableString ss = new ...
- 分布式搜索Elasticsearch——QueryBuilders.matchPhrasePrefixQuery
注:该文项目基础为分布式搜索Elasticsearch——项目过程(一)和分布式搜索Elasticsearch——项目过程(二),项目骨架可至这里下载. ES源代码中对matchPhrasePrefi ...
- 【转】Ubuntu乱码解决方案(全)
转自:http://www.cnblogs.com/end/archive/2011/04/19/2021507.html ubuntu下中文乱码解决方案(全) 1.ibus输入法 Ubuntu 系统 ...
- 使用dreamever去掉文件头部BOM(bom)信息 From 百度经验
本文来此百度经验: 地址为:http://jingyan.baidu.com/article/3f16e003c3dc172591c103e6.html OM主要处理浏览器窗口与框架,但事实上,浏览器 ...
- 【HDOJ】3727 Jewel
静态区间第K大值.主席树和划分树都可解. /* 3727 */ #include <iostream> #include <sstream> #include <stri ...
- C语言计算程序运行时间
#include<stdio.h>#include<stdlib.h> #include "time.h" int main( void ) { ...
- poj 2506 Tiling(递推 大数)
题目:http://poj.org/problem?id=2506 题解:f[n]=f[n-2]*2+f[n-1],主要是大数的相加; 以前做过了的 #include<stdio.h> # ...
- 函数fsp_alloc_seg_inode_page
分配一个新的inode page /**********************************************************************//** Allocat ...