Combinations

Medium

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],
]

Solution

Backtracking

backtracking模板的应用，这里的条件备选方案不重复，排过序（1,2,3,... n）

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> ans = new ArrayList<>();
        if (n == 0 || k == 0) {
            return ans;
        }
        List<Integer> combination = new ArrayList<Integer>();
        combineHelper(1, n, k, combination, ans);

        return ans;
    }

    private void combineHelper(int start, int n, int k, List<Integer> combination, List<List<Integer>> ans) {
        if (combination.size() == k) {
            ans.add(new ArrayList<Integer>(combination));
            return;
        }
        for (int i = start; i <= n; i++) {
            combination.add(i);
            combineHelper(i + 1, n, k, combination, ans);
            combination.remove(combination.size() - 1);
        }
    }
}

DFS

排列类搜索的递归方式，每一次递归决定Index选或者不选

public class Solution {
    /**
     * @param n: Given the range of numbers
     * @param k: Given the numbers of combinations
     * @return: All the combinations of k numbers out of 1..n
     */
    public List<List<Integer>> combine(int n, int k) {
        // write your code here
        List<List<Integer>> result = new ArrayList<>();

        dfs(n, 1, k, new ArrayList<>(), result);

        return result;
    }

    private void dfs(int n, int index, int k, List<Integer> list, List<List<Integer>> result) {
        if (list.size() == k) {
            result.add(new ArrayList<>(list));
            return;
        }

        if (index > n) {
            return;
        }

        list.add(index);
        dfs(n, index + 1, k, list, result);
        list.remove(list.size() - 1);
        dfs(n, index + 1, k, list, result);
    }
}