两道题,都在看视频之前,凭自己的本事写出来了,我哭死

216. 组合总和 III

LinkedList<Integer> path = new LinkedList<>();
    List<List<Integer>> result = new ArrayList<>();

    /**
     * @param k 规模 k 个数
     * @param n 目标 和为 n
     * @return
     */
    public List<List<Integer>> combinationSum3(int k, int n) {
        backTracing3(k,n,0,0);
        return result;
    }

    public void backTracing3(int k, int n, int startIndex,int sum) {
        if(sum > n){
            return;
        }
        if(path.size() > k){
            return;
        }
        if (path.size() == k) {
            if (sum == n) {
                result.add(new ArrayList<>(path));
            }
            return;
        }
        for (int i = startIndex; i < 9; i++) {
            path.add(i + 1);
            sum += i + 1;
            backTracing3(k, n, i + 1,sum);
            sum -= i + 1;
            path.removeLast();
        }
    }

17. 电话号码的字母组合

class Solution {


    String[] list = {"","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    // list  存放数字对应路径

    StringBuilder sb = new StringBuilder();
    List<String> res = new ArrayList<>();
    public List<String> letterCombinations(String digits) {
        if("".equals(digits)){
            return res;
        }
        backTracing(digits.length(),digits,0);
        return res;
    }

    public void backTracing(int length,String digits,int index){
        if(sb.length() == length){
            res.add(String.valueOf(sb));
            return;
        }
        String str = list[ digits.charAt(index) - '1' ];
        for(int i = 0 ; i < str.length() ; i++){
            sb.append(str.charAt(i));
            index++;
            backTracing(length,digits,index);
            index--;
            sb.deleteCharAt(sb.length()-1);
        }

    }
    
    
}