算法修炼Day25|● 216.组合总和III ● 17.电话号码的字母组合LeetCode:216. 组合总和 III

1.思路确定递归方法及其参数（传参的数量和最终实现目的要明确。）；确定终止条件；确定单层递归逻辑（需要回溯）。

2.代码实现

class Solution {
    List<List<Integer>> resList = new ArrayList<>();
    List<Integer> path =  new ArrayList<>();
    // int sum;
    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 0, 1);
        return resList;
    }

    void backtracking(int k, int n, int sum, int startIndex) { // sum和startIndex 很关键
        if (path.size() == k) {
            if (sum == n) {
                resList.add(new ArrayList<>(path));
            }
            return;
        }

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

3.复杂度分析时间复杂度：O(k*(k-1)*···*m).

空间复杂度：O(n).

LeetCode:

1.思路

确定递归函数及其参数；确定单层递归的逻辑；确定终止条件； 2.代码实现

class Solution {
    List<String> list = new ArrayList<>();
    StringBuilder sb = new StringBuilder();

    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) {
            return list;
        }
        String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        backtracking(digits, numString, 0);
        return list;
    }
    void backtracking(String digits, String[] numString, int index) {
        if (index == digits.length()) {
            list.add(sb.toString());
            return;
        }
        String str = numString[digits.charAt(index) - '0'];
        for (int i = 0; i < str.length(); i++) {
            sb.append(str.charAt(i));
            backtracking(digits, numString, index + 1);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}

3.复杂度分析时间复杂度：O(m).

空间复杂度：O(n).