brute-force/combinations
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2023-12-01
Combinations
描述
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example, If n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
递归
// Combinations
// 深搜，递归
// 时间复杂度O(n!)，空间复杂度O(n)
public class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
dfs(n, k, 1, 0, path, result);
return result;
}
// start，开始的数, cur，已经选择的数目
private static void dfs(int n, int k, int start, int cur,
List<Integer> path, List<List<Integer>> result) {
if (cur == k) {
result.add(new ArrayList<>(path));
}
for (int i = start; i <= n; ++i) {
path.add(i);
dfs(n, k, i + 1, cur + 1, path, result);
path.remove(path.size() - 1);
}
}
}
// Combinations
// 深搜，递归
// 时间复杂度O(n!)，空间复杂度O(n)
class Solution {
public:
vector<vector<int> > combine(int n, int k) {
vector<vector<int> > result;
vector<int> path;
dfs(n, k, 1, 0, path, result);
return result;
}
private:
// start，开始的数, cur，已经选择的数目
static void dfs(int n, int k, int start, int cur,
vector<int> &path, vector<vector<int> > &result) {
if (cur == k) {
result.push_back(path);
}
for (int i = start; i <= n; ++i) {
path.push_back(i);
dfs(n, k, i + 1, cur + 1, path, result);
path.pop_back();
}
}
};