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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

}

```