第24，25天：基础回溯，简单！

最新推荐文章于 2025-12-06 07:43:37 发布

原创最新推荐文章于 2025-12-06 07:43:37 发布 · 69 阅读

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#leetcode #算法 #java

这些回溯题寒假刚刚刷过，所以快速练练手过一遍～

1.Combinations

class Solution {
    private void backtracking(List<List<Integer>> res, List<Integer> tmp, int start, int n, int k){
        if(tmp.size() == k){
            res.add(new ArrayList<Integer>(tmp));
            return;
        }

        for(int i = start; i <= n; ++i){
            tmp.add(i);
            backtracking(res, tmp, i+1, n, k);
            tmp.remove(tmp.size()-1);
        }
    }
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();

        backtracking(res, tmp, 1, n, k);
        return res;
    }
}

2.Combination Sum III

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> tmp = new ArrayList<>();
    int sum = 0;

    private void backtracking(int k, int n, int start){
        if(sum == n && tmp.size() == k){
            res.add(new ArrayList<Integer>(tmp));
            return;
        }

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

    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);

        return res;
    }
}

进行两个剪枝：

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> tmp = new ArrayList<>();
    int sum = 0;

    private void backtracking(int k, int n, int start){
        if(tmp.size() == k){
            if(sum == n) res.add(new ArrayList<Integer>(tmp));
            return;
        }
        if(sum >= n) return;

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

    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);

        return res;
    }
}

3.Letter Combinations of a Phone Number

只是把回溯变成了string而已，就这？

class Solution {
    List<String> res = new ArrayList<>();
    String arr[] = {"", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    private void backtracking(String digits, int start, String tmp){
        if(tmp.length() == digits.length()){
            String copy = tmp;
            res.add(copy);
            return;
        }
        int used = Integer.parseInt(digits.substring(start, start+1))-1;
        for(int i = 0; i < arr[used].length(); ++i){
            tmp += arr[used].substring(i, i+1);
            backtracking(digits, start+1, tmp);
            tmp = tmp.substring(0, tmp.length()-1);
        }

    }

    public List<String> letterCombinations(String digits) {
        if(digits.equals("")) return res;
        backtracking(digits, 0, "");
        return res;
    }
}