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M
1526624015
tags: DFS, Backtracking, Combination
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
#### DFS, Backtracking
- for loop, recursive (dfs).
- 每个item用一次, 下一个level dfs(index + 1)
- Combination DFS. 画个图想想. 每次从1~n里面pick一个数字i
- 因为下一层不能重新回去 [0~i]选，所以下一层recursive要从i+1开始选。
```
/*
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],
]
*/
/*
dfs, for loop, track index, track size of list.
once size met, add to result, also no need to further dfs.
dfs: result,list, index, k, n
*/
class Solution {
public List<List<Integer>> combine (int n, int k) {
List<List<Integer>> result = new ArrayList<>();
// check edge case
if (k <= 0 || n <= 0) {
return result;
}
// init result, dfs
dfs(result, new ArrayList<>(), 1, k , n);
return result;
}
private void dfs(List<List<Integer>> result, List<Integer> list, int index, int k, int n) {
// for loop
// check size, dfs with i + 1
for (int i = index; i <= n; i++) {
list.add(i);
// add to result
if (list.size() == k) {
result.add(new ArrayList<>(list));
list.remove(list.size() - 1);
continue;
}
dfs(result, list, i + 1, k, n);
list.remove(list.size() - 1);
}
}
}
/*
Thinking process:
Use a helper method to perform recursive backtracking:add an element to next-level recursive call,
and remote the entry after the recursive call.
Note: When 'new' something, cannot use 'List' because it's a abstract class. Need to new 'ArrayList'
*/
public class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> rst = new ArrayList<List<Integer>>();
if (n <= 0 || k <= 0) {
return rst;
}
List<Integer> solution = new ArrayList<Integer>();
helper(rst, solution, n, k, 1);// Start == 1 because we want 1 ~ n in this problem
return rst;
}
public void helper(List<List<Integer>> rst,
List<Integer> solution, int n, int k, int start) {
if (solution.size() == k) {
rst.add(new ArrayList(solution));
return;
}
for (int i = start; i <= n; i++) {// <=n because we want 1 ~ n in this problem
solution.add(i);
helper(rst, solution, n, k, i + 1);
solution.remove(solution.size() - 1); //Back-track
}
}
}
```