Leetcode -- Day 61

Combination

Question 1

Combinations

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],
]
It is also a backstracking tagged problem.

    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> rs = new ArrayList<List<Integer>>();
        dfs(rs, n, 1, new ArrayList<Integer>(), k);
        return rs;
    }
    private void dfs(List<List<Integer>> rs, int n, int start, List<Integer> path, int k ){
        //the terminate condition, return when every process is finished 
        if(k==0){
            rs.add(new ArrayList<Integer>(path));
            return;
        }
        //start from 1 to n (i.e. 1,2,3, ... ,n)
        for(int i = start; i<=n; i++){
            path.add(i);
            dfs(rs, n, i+1, path, k-1);
            path.remove(path.size() - 1);
        }

    }

Question 2

Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

public class Solution {
    
    private String[] button;
    public List<String> letterCombinations(String digits) {
        button = new String[10];
        button[0] = "";
        button[1] = "";
        button[2] = "abc";
        button[3] = "def";
        button[4] = "ghi";
        button[5] = "jkl";
        button[6] = "mno";
        button[7] = "pqrs";
        button[8] = "tuv";
        button[9] = "wxyz";
        
        List<String> result = new ArrayList<String>();
        if (digits == null || digits.length() < 1){
            return result;
        }
        dfs(digits, 0, "", result);
        return result;
    }
    
    public void dfs(String digits, int start, String path, List<String> result){
        if (start >= digits.length()){
            result.add(path);
            return;
        }
        String content = button[digits.charAt(start)-'0'];
        for (int i = 0; i < content.length(); i ++){
            dfs(digits, start + 1, path+content.charAt(i), result);
        }
    }
}