day25 打卡216.组合总和III 17.电话号码的字母组合

216.组合总和III

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

    public List<List<Integer>> combinationSum3(int k, int n) {
        backTracking(k, n, 1, 0);
        return result;
    }

    private void backTracking(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 LinkedList<>(path));
            }
        }
        

        for (int i=startIndex; i<=9 ; i++) {
            path.add(i);
            sum += i;
            backTracking(k, n, i+1, sum);
            sum -= i;
            path.removeLast();
        }
    }
}

17.电话号码的字母组合

17题目链接

class Solution {
    //设置全局列表存储最后的结果
    List<String> list = new ArrayList<>();

    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) return list;
        backTracking(digits, 0);
        return list;
    }

    //每次迭代获取一个字符串，所以会设计大量的字符串拼接，所以这里选择更为高效的 StringBuild
    StringBuilder sb = new StringBuilder();
    String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

    private void backTracking(String digits, int index) {
        if (index == digits.length()) {
            list.add(sb.toString());
            return;
        }

        Integer num = Integer.parseInt(digits.charAt(index)+"");
        String str = numString[num];
        for (int i = 0 ; i<str.length() ; i++) {
            sb.append(str.charAt(i));
            backTracking(digits, index+1);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}