1215. Algorithm - Combination and PermutationCombination, Permutation, and Subset
Implement DFS for combination, permutation and subset.
1. Combination
1.1 Description
Given two integers n
and k
, return all possible combinations of k
numbers out of 1 … n.
Example:
Input: n = 4, k = 2
Output:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
1.2 DFS Implementation
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> ans = new ArrayList<>();
if (n <= 0 || k <= 0 || n < k) {
return ans;
}
List<Integer> list = new ArrayList<Integer>();
dfs(n, k, 1, list, ans);
return ans;
}
private void dfs(int n, int k, int pos, List<Integer> list, List<List<Integer>> ans) {
if (list.size() == k) {
ans.add(new ArrayList<Integer>(list));
return;
}
for(int i = pos; i <= n; i++) {
list.add(i);
dfs(n, k, i + 1, list, ans);
list.remove(list.size() - 1);
}
}
2. Permutation
2.1 Description
Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]
2.2 DFS
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
if (nums == null || nums.length == 0) {
return ans;
}
boolean[] visited = new boolean[nums.length];
dfs(nums, visited, new ArrayList<>(), ans);
return ans;
}
private void dfs(int[] nums, boolean[] visited, List<Integer> list, List<List<Integer>> ans) {
if (list.size() == nums.length) {
ans.add(new ArrayList<>(list));
return;
}
for (int i = 0; i < nums.length; i++) {
if (visited[i]) {
continue;
}
visited[i] = true;
list.add(nums[i]);
dfs(nums, visited, list, ans);
list.remove(list.size() - 1);
visited[i] = false;
}
}
3. Subsets
3.1 Description
Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:
Input: nums = [1,2,3]
Output:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
3.2 DFS
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
if (nums == null || nums.length == 0) {
return ans;
}
//Arrays.sort(nums); // not necessary, just for unit test
dfs(nums, 0, new ArrayList<>(), ans);
return ans;
}
private void dfs(int[] nums, int pos, List<Integer> list, List<List<Integer>> ans) {
ans.add(new ArrayList<>(list));
for (int i = pos; i < nums.length; i++) {
list.add(nums[i]);
dfs(nums, i + 1, list, ans);
list.remove(list.size() - 1);
}
}
4. Summary
- Combinations: need to use
pos
, no for loop in the main function. - Permutations: no need to use
pos
, usevisited
array to store which numbers are used, no for loop in the main function. - Subsets: need to use
pos
, no for loop in the main function.
5. Classic Problems
- LeetCode 77 - Combinations
- LeetCode 39 - Combination Sum
- LeetCode 40 - Combination Sum II
- LeetCode 216 - Combination Sum III
- LeetCode 46 - Permutations
- LeetCode 78. Subsets