LeetCode_Combination Sum
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times. Note: All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, � , ak) must be in non-descending order. (ie, a1 ? a2 ? � ? ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
class Solution {
public:
void DFS(vector<int> &candidates, int target, int start, int sum, vector<int> &tp){
if(sum == target){
res.push_back(tp);
return;
}
for(int i = start; i< candidates.size(); ++i){
if(candidates[i] + sum <= target){
tp.push_back(candidates[i]);
DFS(candidates, target, i, sum+candidates[i], tp);
tp.pop_back();
}
}
}
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
res.clear();
int len = candidates.size();
if(len < 1 || target <1) return res;
sort(candidates.begin(), candidates.end());
vector<int> tp;
DFS(candidates, target, 0, 0, tp);
return res;
}
private:
vector<vector<int>> res;
};
LeetCode_Combination Sum的更多相关文章
- LeetCode_Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in ...
- LeetCode - Two Sum
Two Sum 題目連結 官網題目說明: 解法: 從給定的一組值內找出第一組兩數相加剛好等於給定的目標值,暴力解很簡單(只會這樣= =),兩個迴圈,只要找到相加的值就跳出. /// <summa ...
- Leetcode 笔记 113 - Path Sum II
题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each ...
- Leetcode 笔记 112 - Path Sum
题目链接:Path Sum | LeetCode OJ Given a binary tree and a sum, determine if the tree has a root-to-leaf ...
- POJ 2739. Sum of Consecutive Prime Numbers
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20050 ...
- BZOJ 3944 Sum
题目链接:Sum 嗯--不要在意--我发这篇博客只是为了保存一下杜教筛的板子的-- 你说你不会杜教筛?有一篇博客写的很好,看完应该就会了-- 这道题就是杜教筛板子题,也没什么好讲的-- 下面贴代码(不 ...
- [LeetCode] Path Sum III 二叉树的路径和之三
You are given a binary tree in which each node contains an integer value. Find the number of paths t ...
- [LeetCode] Partition Equal Subset Sum 相同子集和分割
Given a non-empty array containing only positive integers, find if the array can be partitioned into ...
- [LeetCode] Split Array Largest Sum 分割数组的最大值
Given an array which consists of non-negative integers and an integer m, you can split the array int ...
随机推荐
- 几种开源SIP协议栈对比OPAL,VOCAL,sipX,ReSIProcate,oSIP
随着VoIP和NGN技术的发展,H.323时代即将过渡到SIP时代,在H.323的开源协议栈中,Openh323占统治地位,它把一个复杂而又先进 的H.323协议栈展现在普通程序员的眼前,为H.323 ...
- Apache HttpClient组件封装工具类
package com.mengyao.spider.utils; import java.util.ArrayList;import java.util.HashMap;import java.ut ...
- java中protect属性用法总结
测试代码: pojo类: package com.lky.h1; public class Base { private Integer id; protected String name; publ ...
- brent ozar的sqlserver dba训练课程翻译——第二章:手动恢复数据库
备份的唯一原因 备份的唯一原因是我们可以还原 当我第一次成为sqlserver数据库管理员,只要备份工作都能成功运行,我就会觉得一切都很好.我会查看sqlserver代理,保证那些作业都在运行,然 ...
- 对Ul下的li标签执行点击事件——如何获取你所点击的标签
问题所来:做项目时,一般的数据都是用循环动态加载出来的,结构都是一样的,只是绑定的值不同,如何对相同的标签做处理的问题就来了. 例如:点谁就显示谁的数值 <ul > <li id=& ...
- Hadoop 类Grep源代码注释
/** * Licensed to the Apache Software Foundation (ASF) under one * or more contributor license agree ...
- Linux下基于源代码方式安装MySQL 5.6
MySQL为开源数据库,因此能够基于源代码实现安装.基于源代码安装有很多其它的灵活性. 也就是说我们能够针对自己的硬件平台选用合适的编译器来优化编译后的二进制代码.依据不同的软件平台环境调整相关的编译 ...
- source insight3.5中文乱码解决方案
source insight3.5中文乱码,网上看别人说改变宽字体.宋体等方法都不起效.根本原因是,source insight 3.5 不支持Unicode编码,所以导致中文的乱码,将文件转为gb2 ...
- C#。1 数据类型,常量变量,类型转换
C#. 一.数据类型 1,字符串类型(string) .放入一串字符串,需要用""引起来. 列如: string a ="999"; 2,整型 (int). ...
- Android(通用机能)
数据存储 本地数据存在都是私有化的. 共享方法1是构造数据源组件.方法2将数据放入扩展存储设备. Mashup 服务组件默认没有运行在独立进程或线程中,因此费时操作一般需要起线程.可配置指定新进程. ...