notes/OJ notes/pages/Leetcode Combinations.md
2022-07-20 14:25:33 +08:00

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# Leetcode Combinations
#### 2022-07-19 20:09
> ##### Algorithms:
> #algorithm #backtrack
> ##### Difficulty:
> #coding_problem #difficulty-medium
> ##### Additional tags:
> #leetcode #CS_list_need_understanding
> ##### Revisions:
> N/A
##### Related topics:
```expander
tag:#backtrack
```
##### Links:
- [Link to problem](https://leetcode.com/problems/combinations/)
- [Explanation](https://leetcode.com/problems/combinations/discuss/844096/Backtracking-cheatsheet-%2B-simple-solution)
___
### Problem
Given two integers `n` and `k`, return _all possible combinations of_ `k` _numbers out of the range_ `[1, n]`.
You may return the answer in **any order**.
#### Examples
**Example 1:**
```
**Input:** n = 4, k = 2
**Output:**
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
```
**Example 2:**
```
**Input:** n = 1, k = 1
**Output:** [[1]]
```
#### Constraints
**Constraints:**
- `1 <= n <= 20`
- `1 <= k <= n`
### Thoughts
> [!summary]
> This is a #backtrack problem
The link I mentioned on the links section already have a good
explanation on the topic.
In my eyes, backtracking means DFS-like recursion and searching in solving problems.
First, it add a option in the combinations, then backtrack it
And when it's done, it gets poped out and try another solution, and backtrack.
### Solution
Backtracking
```cpp
class Solution {
void backtrack(vector<vector<int>> &ans, vector<int> &combs, int n, int nex,
int k) {
if (k == 0) {
ans.push_back(combs);
} else {
// try different solutions
for (int i = nex; i <= n; i++) {
combs.push_back(i);
backtrack(ans, combs, n, i + 1, k - 1);
combs.pop_back();
}
}
}
public:
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> ans = {};
vector<int> combs = {};
backtrack(ans, combs, n, 1, k);
return ans;
}
};
```