Given two integers n and k, return all possible combinations ofknumbers chosen from the range[1, n].
You may return the answer in any order.
Example 1:
Input: n = 4, k = 2
Output: [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Explanation: There are 4 choose 2 = 6 total combinations.
Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination.
Example 2:
Input: n = 1, k = 1
Output: [[1]]
Explanation: There is 1 choose 1 = 1 total combination.
Constraints:
1 <= n <= 20
1 <= k <= n
Solutions
Solution 1: Backtracking (Two Ways)
We design a function $dfs(i)$, which represents starting the search from number $i$, with the current search path as $t$, and the answer as $ans$.
The execution logic of the function $dfs(i)$ is as follows:
If the length of the current search path $t$ equals $k$, then add the current search path to the answer and return.
If $i \gt n$, it means the search has ended, return.
Otherwise, we can choose to add the number $i$ to the search path $t$, and then continue the search, i.e., execute $dfs(i + 1)$, and then remove the number $i$ from the search path $t$; or we do not add the number $i$ to the search path $t$, and directly execute $dfs(i + 1)$.
The above method is actually enumerating whether to select the current number or not, and then recursively searching the next number. We can also enumerate the next number $j$ to be selected, where $i \leq j \leq n$. If the next number to be selected is $j$, then we add the number $j$ to the search path $t$, and then continue the search, i.e., execute $dfs(j + 1)$, and then remove the number $j$ from the search path $t$.
In the main function, we start the search from number $1$, i.e., execute $dfs(1)$.
The time complexity is $(C_n^k \times k)$, and the space complexity is $O(k)$. Here, $C_n^k$ represents the combination number.