69 lines
2.4 KiB
Markdown
69 lines
2.4 KiB
Markdown
|
---
|
|||
|
|
id: "202203103"
|
|||
|
|
date: "2022/03/10 22:15"
|
|||
|
|
title: "Q216 组合总和3(combination-sumⅢ)"
|
|||
|
|
tags: ["java", "leetcode", "dfs"]
|
|||
|
|
index_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg?imageView2/2/w/200
|
|||
|
|
banner_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg
|
|||
|
|
categories:
|
|||
|
|
- "算法"
|
|||
|
|
- "leetcode刷题"
|
|||
|
|
---
|
|||
|
|
|
|||
|
|
<span id="blogIdSpan" style="display:none">202203103</span>
|
|||
|
|
|
|||
|
|
## 解析思路
|
|||
|
|
|
|||
|
|
leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/combination-sum-iii/)。
|
|||
|
|
|
|||
|
|
本题属于[组合排序 2](https://blog.fleyx.com/blog/detail/202203102/)的进阶题型。建议先看[上一篇](https://blog.fleyx.com/blog/detail/202203102/)的解析。
|
|||
|
|
|
|||
|
|
本题跟上一题区别只有一点是**选择的数字不能重复**,**同时对选择的数量有限制**如何实现呢？
|
|||
|
|
|
|||
|
|
很简单，只需要
|
|||
|
|
|
|||
|
|
- 将传入的 index 设置成 index+1，实现不重复
|
|||
|
|
- ~~同时 for 循环时每次都循环到下一个不同的元素上~~由于本来提供的数字就不重复，因此可以去掉这一条
|
|||
|
|
- 循环和返回结果是都判断下元素数量
|
|||
|
|
|
|||
|
|
代码如下：
|
|||
|
|
|
|||
|
|
## 代码
|
|||
|
|
|
|||
|
|
```java
|
|||
|
|
public class Q216 {
|
|||
|
|
public List<List<Integer>> combinationSum3(int k, int n) {
|
|||
|
|
int[] arr = {1, 2, 3, 4, 5, 6, 7, 8, 9};
|
|||
|
|
List<List<Integer>> res = new ArrayList<>();
|
|||
|
|
dfs(arr, n, k, 0, new Stack<>(), res);
|
|||
|
|
return res;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
private void dfs(int[] candidates, int n, int k, int index, Stack<Integer> temp, List<List<Integer>> res) {
|
|||
|
|
if (n == 0) {
|
|||
|
|
//说明找到一个结果序列
|
|||
|
|
if (temp.size() == k) {
|
|||
|
|
//只有刚好k个数才是结果
|
|||
|
|
res.add(new ArrayList<>(temp));
|
|||
|
|
}
|
|||
|
|
return;
|
|||
|
|
}
|
|||
|
|
for (int i = index; i < candidates.length; i++) {
|
|||
|
|
if (candidates[i] > n || temp.size() == k) {
|
|||
|
|
//前面已经排序过，所以在这里可以进行剪枝操作，如果candidates[index]都小于target了，那就不需要比较后面的了，肯定不满足要求
|
|||
|
|
//另外只能使用k个数，所以当temp的size为k时说明不能在装了，本条路结束
|
|||
|
|
return;
|
|||
|
|
}
|
|||
|
|
temp.push(candidates[i]);
|
|||
|
|
dfs(candidates, n - candidates[i], k, i + 1, temp, res);
|
|||
|
|
temp.pop();
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
public static void main(String[] args) {
|
|||
|
|
new Q216().combinationSum3(3, 9).forEach(System.out::println);
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
}
|
|||
|
|
```
|