Day 25 | 216. Combination Sum III | 17. Letter Combinations of a Phone Number

LeetCode算法题解：递归与回溯法

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文章介绍了在LeetCode上的一系列编程题目，涉及二叉树的遍历、链表操作、数组问题以及字符串处理等，主要使用了回溯算法来解决，如组合求和、电话号码字母组合等。

LeetCode 216. Combination Sum III

Question Link

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

    void backTracking(int startIndex, int k, int n){
        // pruning
        if(sum > n)
            return;
        if(path.size() == k && sum == n){
            result.add(new ArrayList<>(path));
            return;
        }    
        for(int i = startIndex; i <= 9 - (k - path.size()) + 1; i++){
            path.add(i);
            sum+=i;
            backTracking(i+1, k, n);
            path.remove(path.size() - 1);
            sum-=i;
        }
    }
}

The number of nodes found: path.size()
The number of nodes to find: k - path.size()
The minimum length of [x, 9] should be k - path.size() for continued searching. We can get equation 9 - x + 1 = k - path.size()
Solve the equation to get x = n - (k - path.size()) + 1

LeetCode 17. Letter Combinations of a Phone Number

Question Link

class Solution {
    String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    List<String> result = new ArrayList<>();
    StringBuilder path = new StringBuilder();

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

    void backTracking(int index, String digits){
        if(index == digits.length()){
            result.add(path.toString());
            return;
        }
        String str = numString[digits.charAt(index) - '0'];
        for(int i = 0; i < str.length(); i++){
            path.append(str.charAt(i));
            backTracking(index + 1, digits);
            path.deleteCharAt(path.length()-1);
        }
    }
}