LC 377. Combination Sum IV
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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
Runtime: 2 ms, faster than 80.43% of Java online submissions for Combination Sum IV.
class Solution {
public int combinationSum4(int[] nums, int target) {
int[] dp = new int[target+1];
dp[0] = 1;
for(int i=0; i<dp.length; i++){
for(int j=0; j<nums.length; j++){
if(i >= nums[j]){
dp[i] += dp[i - nums[j]];
}
}
}
return dp[target];
}
}
LC 377. Combination Sum IV的更多相关文章
- 39. Combination Sum + 40. Combination Sum II + 216. Combination Sum III + 377. Combination Sum IV
▶ 给定一个数组 和一个目标值.从该数组中选出若干项(项数不定),使他们的和等于目标值. ▶ 36. 数组元素无重复 ● 代码,初版,19 ms .从底向上的动态规划,但是转移方程比较智障(将待求数分 ...
- [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 ...
- [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 ...
- 377. Combination Sum IV 返回符合目标和的组数
[抄题]: Given an integer array with all positive numbers and no duplicates, find the number of possibl ...
- 377. Combination Sum IV
问题 Given an integer array with all positive numbers and no duplicates, find the number of possible c ...
- Leetcode 377. Combination Sum IV
Given an integer array with all positive numbers and no duplicates, find the number of possible comb ...
- 377. Combination Sum IV——DP本质:针对结果的迭代,dp[ans] <= dp[ans-i] & dp[i] 找三者关系 思考问题的维度+1,除了数据集迭代还有考虑结果
Given an integer array with all positive numbers and no duplicates, find the number of possible comb ...
- 377. Combination Sum IV 70. Climbing Stairs
back function (return number) remember the structure class Solution { int res = 0; //List<List< ...
- 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 ...
随机推荐
- go语言入门(6)复合类型
1,分类 2,指针 指针是一个代表着某个内存地址的值.这个内存地址往往是在内存中存储的另一个变量的值的起始位置. 1)基本操作 Go语言虽然保留了指针,但与其它编程语言不同的是: 默认值 nil,没有 ...
- InnoDB意向锁的作用
https://www.zhihu.com/question/51513268?sort=created IX,IS是表级锁,不会和行级的X,S锁发生冲突.只会和表级的X,S发生冲突. 意向锁是一种快 ...
- 【Java并发】锁机制
一.重入锁 二.读写锁 三.悲观锁.乐观锁 3.1 悲观锁 3.2 乐观锁 3.3 CAS操作方式 3.4 CAS算法理解 3.5 CAS(乐观锁算法) 3.6 CAS缺点 四.原子类 4.1 概述 ...
- Go语言根据数据表自动生成model以及controller代码
手写model的用法请参考: https://www.jianshu.com/p/f5784b8c00d0 这里仅说明自动生成model文件的过程 bee generate appcode -tabl ...
- OpenCV入门学习资料汇总
OpenCV学习文档资料 OpenCV学习:1)OpenCV中文网站——http://wiki.opencv.org.cn/index.php/%E9%A6%96%E9%A1%B5 2)python实 ...
- davinci 删除路线站点关系
删除路线站点关系 DELETE FROM tb_station_info_draw WHERE id in (SELECT stationId FROM tb_road_station_relatio ...
- Matlab---三维视图的自动旋转
Matlab---三维视图的自动旋转 这里给出一种格式说明: % figure % plot3(...); % xlabel('X轴'); % ylabel('Y轴'); % zlabel('Z轴') ...
- pyltp安装
第一步:下载wheel文件 第二步:进入该文件的文件夹 第三步:pip install wheel文件名 注意:python的安装版本必须和pyltp的版本相同,我这版本都是pyhton3.6.之前p ...
- JS中BOM和DOM之间的关系
一.Javascript组成JavaScript的实现包括以下3个部分:1.核心(ECMAScript):描述了JS的语法和基本对象.2.文档对象模型 (DOM):处理网页内容的方法和接口.3.浏览器 ...
- 端口与服务-ftp服务
端口与服务-ftp服务 1概述 1.1.从先知和乌云上爬取端口历史漏洞报告,总结报告 1.2.全面总结,出具一个表格之类的汇总表 2.ftp # -*- coding: utf-8 -*- impor ...