Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

这道题题是组合之和系列的第三道题,跟之前两道 Combination SumCombination Sum II 都不太一样,那两道的联系比较紧密,变化不大,而这道跟它们最显著的不同就是这道题的个数是固定的,为k。个人认为这道题跟那道 Combinations 更相似一些,但是那道题只是排序,对k个数字之和又没有要求。所以实际上这道题是它们的综合体,两者杂糅到一起就是这道题的解法了,n是k个数字之和,如果n小于0,则直接返回,如果n正好等于0,而且此时out中数字的个数正好为k,说明此时是一个正确解,将其存入结果res中,具体实现参见代码入下:

class Solution {
public:
vector<vector<int> > combinationSum3(int k, int n) {
vector<vector<int> > res;
vector<int> out;
combinationSum3DFS(k, n, , out, res);
return res;
}
void combinationSum3DFS(int k, int n, int level, vector<int> &out, vector<vector<int> > &res) {
if (n < ) return;
if (n == && out.size() == k) res.push_back(out);
for (int i = level; i <= ; ++i) {
out.push_back(i);
combinationSum3DFS(k, n - i, i + , out, res);
out.pop_back();
}
}
};

类似题目:

Combination Sum IV

Combination Sum II

Combination Sum

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] Combination Sum III 组合之和之三的更多相关文章

  1. [LeetCode] 216. Combination Sum III 组合之和 III

    Find all possible combinations of k numbers that add up to a number n, given that only numbers from ...

  2. [LeetCode] Combination Sum IV 组合之和之四

    Given an integer array with all positive numbers and no duplicates, find the number of possible comb ...

  3. [LeetCode] Combination Sum II 组合之和之二

    Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in ...

  4. [Leetcode] combination sum ii 组合之和

    Given a collection of candidate numbers ( C ) and a target number ( T), find all unique combinations ...

  5. [LeetCode] 377. Combination Sum IV 组合之和 IV

    Given an integer array with all positive numbers and no duplicates, find the number of possible comb ...

  6. [LeetCode] 377. Combination Sum IV 组合之和之四

    Given an integer array with all positive numbers and no duplicates, find the number of possible comb ...

  7. [LeetCode] 40. Combination Sum II 组合之和之二

    Given a collection of candidate numbers (candidates) and a target number (target), find all unique c ...

  8. [leetcode]40. Combination Sum II组合之和之二

    Given a collection of candidate numbers (candidates) and a target number (target), find all unique c ...

  9. [LeetCode] 40. Combination Sum II 组合之和 II

    Given a collection of candidate numbers (candidates) and a target number (target), find all unique c ...

随机推荐

  1. ASP.NET Core 中文文档 第四章 MVC(01)ASP.NET Core MVC 概览

    原文:Overview of ASP.NET Core MVC 作者:Steve Smith 翻译:张海龙(jiechen) 校对:高嵩 ASP.NET Core MVC 是使用模型-视图-控制器(M ...

  2. Windows环境下vscode-go安装笔记

    一.介绍 对于Visual Studio Code开发工具,有一款优秀的GoLang插件,它的主页为:https://github.com/microsoft/vscode-go 这款插件的特性包括: ...

  3. Apworks框架实战(四):使用Visual Studio开发面向经典分层架构的应用程序:从EasyMemo案例开始

    时隔一年,继续我们的Apworks框架之旅.在接下来的文章中,我将逐渐向大家介绍如何在Visual Studio中结合Apworks框架,使用ASP.NET Web API和MVC来开发面向经典分层架 ...

  4. Basic Tutorials of Redis(8) -Transaction

    Data play an important part in our project,how can we ensure correctness of the data and prevent the ...

  5. Ionic2系列——使用DeepLinker实现指定页面URL

    Ionic2使用了近似原生App的页面导航方式,并不支持Angular2的路由.这种方式在开发本地App的时候比较方便,但如果要用来开发纯Web页面就有点问题了,这种情况下Angular2的route ...

  6. 【无私分享:ASP.NET CORE 项目实战(第十四章)】图形验证码的实现

    目录索引 [无私分享:ASP.NET CORE 项目实战]目录索引 简介 很长时间没有来更新博客了,一是,最近有些忙,二是,Core也是一直在摸索中,其实已经完成了一个框架了,并且正在准备在生产环境中 ...

  7. C#开发微信门户及应用(20)-微信企业号的菜单管理

    前面几篇陆续介绍了很多微信企业号的相关操作,企业号和公众号一样都可以自定义菜单,因此他们也可以通过API进行菜单的创建.获取列表.删除的操作,因此本篇继续探讨这个主体,介绍企业号的菜单管理操作. 菜单 ...

  8. spring笔记5 spring IOC的基础知识1

    1,ioc的概念 Inverse of control ,控制反转,实际的意义是调用类对接口实现类的依赖,反转给第三方的容器管理,从而实现松散耦合: ioc的实现方式有三种,属性注入,构造函数注入,接 ...

  9. MYSQL 开发技巧

    主要涉及:JOIN .JOIN 更新.GROUP BY HAVING 数据查重/去重 1 INNER JOIN.LEFT JOIN.RIGHT JOIN.FULL JOIN(MySQL 不支持).CR ...

  10. 记录一次bug解决过程:velocity中获取url中的参数

    一.总结 在Webx的Velocity中获取url中参数:$rundata.getRequest().getParameter('userId') 在Webx项目中,防止CSRF攻击(Cross-si ...