Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.
Example:
nums = [1, 2, 3]
target = 4
The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)
Note that different sequences are counted as different combinations.
Therefore the output is 7.

Follow up:
What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

详见:https://leetcode.com/problems/combination-sum-iv/description/

C++:

class Solution {
public:
int combinationSum4(vector<int>& nums, int target) {
vector<int> dp(target+1);
dp[0]=1;
for(int i=1;i<=target;++i)
{
for(int num:nums)
{
if(i>=num)
{
dp[i]+=dp[i-num];
}
}
}
return dp.back();
}
};

参考:https://www.cnblogs.com/grandyang/p/5705750.html

377 Combination Sum IV 组合之和 IV的更多相关文章

  1. [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 ...

  2. [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 ...

  3. [LeetCode] Combination Sum III 组合之和之三

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

  4. [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 ...

  5. [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 ...

  6. LeetCode OJ:Combination Sum II (组合之和 II)

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

  7. [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 ...

  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] combination sum ii 组合之和

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

随机推荐

  1. 网页js粘贴截图

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...

  2. Sum-(最大子序列和)

    http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=652&pid=1003 题目大意: 给你一个序列,你随便找一 ...

  3. HashSet源码分析1

    import java.util.HashSet; import java.util.Iterator; import java.util.Set; public class SetTest { pu ...

  4. JSP中操作Java Beans

    以下内容引用自http://wiki.jikexueyuan.com/project/jsp/beans.html: JavaBean是在编写Java时专门创建的Java类,根据JavaBean AP ...

  5. 怎样删除Tomcat下已经部署的项目

    lz说的是把web项目部署到tomcat之中,要把它删除..很简单,找到webapps文件(tomcat的根目录)下把它删除即可.. 2.Tomcat 6.0\webapps\项目名 只要在把这个目录 ...

  6. 关于warning

    IT男正吸着雪茄,吐着烟圈.他女朋友生气了发飙道:“你没看见包装盒上的警告(Warning)么?吸烟有害健康!” IT男淡定地回答道:“我是程序员.我们不关心警告,只关心错误.”

  7. openstack(2) -------RabbitMQ集群部署

    一.RabbitMQ简介 RabbitMQ属于一个流行的开源消息队列系统.属于AMQP( 高级消息队列协议 ) 标准的一个 实现.是应用层协议的一个开放标准,为面向消息的中间件设计.用于在分布式系统中 ...

  8. 图像处理之基础---用Shader实现的YUV到RGB转换:使用3重纹理实现 .

    上一篇中,我是用一个RGB格式的纹理来存储每一帧的画面,其中纹理为m_FrameWidth * m_FrameHeight大小,这样,在内存中,就必须要先对YUV的数据进行排序,然后才能当做RGB的数 ...

  9. Cocos2d-x项目创建

    以创建HelloWorld项目为例子,执行create_project.py脚本,进入Doc界面输入下面的命令: (1)E:   (切换盘符,因为我的Cocos2d-x源码在E盘,create_pro ...

  10. document.body.className = document.body.className.replace("siteorigin-panels-before-js","");

    document.body.className = document.body.className.replace("siteorigin-panels-before-js",&q ...