77. Combinations变形，还加了字母

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

You may return the answer in any order.

 

Example 1:

Input: n = 4, k = 2
Output:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

Example 2:

Input: n = 1, k = 1
Output: [[1]]

用字母当数字，输出所有combinations。 例子：假设 k = 4, 输出  ["1111", "1112", ... ,"1119", "111A", "111B", ... "ZZZZ"]

最直接的方法就是for四遍吧。

不过肯定要优化：dfs是干嘛的来着？和backtracing有何区别？树的dfs是traverse和dc，图的dfs是traverse和回溯法

肯定是backtracing

    public static List<List<Character>> combine(int k) {
        List<List<Character>> combs = new ArrayList<List<Integer>>();
        List<Character> chars = new ArrayList<Character>{'1', '2', '3'...'z'};

        combine(combs, chars, new ArrayList<Integer>(), 1, k);

        return combs;
    }

    public static void combine(List<List<Character>> combs, List<Character> chars, 
        List<Character> comb, int start, int k) {
        if(k==0) {
            combs.add(new ArrayList<Character>(comb));
            return;
        }

        for(int i=start;i<=chars.size();i++) {
            comb.add(chars[i]);
            combine(combs, comb, i+1, n, k-1);
            comb.remove(comb.size()-1);
        }
    }