代码随想录算法训练营第二十四天 二十五 | 回溯的理论基础，77. 组合 216. 组合总和 III 17. 电话号码的字母组合

77. 组合
https://leetcode.cn/problems/combinations/description/

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

总结：回溯，每次选一个点，for中是这一横层的选点，选完了把下一个点放进去，放完了拿出来。
216. 组合总和 III
https://leetcode.cn/problems/combinations/description/

List<List<Integer>> res;
    List<Integer> path;
    public List<List<Integer>> combinationSum3(int k, int n) {
        res = new ArrayList<>();
        path = new ArrayList<>();
        backtracking(k,n,1);
        return res;
    }
    public void backtracking(int k,int n,int start){
        if (path.size() > k) return;//这句话就是剪枝优化
        if (path.size() == k){
            int count = 0;
            for (Integer integer : path) {
                count += integer;
            }
            if (count == n){
                res.add(new ArrayList<>(path));
                return;
            }else {
                return;
            }
        }
        for (int i = start; i <= 9; i++) {
            path.add(i);
            backtracking(k,n,i+1);
            path.remove(path.size() - 1);
        }
    }

总结：回溯，每次长度够了就看值是否符合，长度超了就直接return
17. 电话号码的字母组合
https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/

List<String> res;
    StringBuilder path;
    HashMap<Integer,String> map;
    int digitsLen;
    public List<String> letterCombinations(String digits) {
        res = new ArrayList<>();
        path = new StringBuilder();
        map = new HashMap<>();
        if (digits.isEmpty()) return res;
        map.put(2,"abc");
        map.put(3,"def");
        map.put(4,"ghi");
        map.put(5,"jkl");
        map.put(6,"mno");
        map.put(7,"pqrs");
        map.put(8,"tuv");
        map.put(9,"wxyz");
        digitsLen = digits.length();
        backtracking(0,digits);
        return res;
    }
    public void backtracking(int num,String digits){
        if (path.length() == digitsLen){
            res.add(path.toString());
            return;
        }
        int key = digits.charAt(num) - '0';
        String str = map.get(key);
        for (int i = 0; i < str.length(); i++) {
            path.append(str.charAt(i));
            backtracking(num+1,digits);
            path.deleteCharAt(path.length()-1);
        }
    }

总结：用回溯，先把每个数字和对应的字符串存到map中，再去遍历字符串的每个位置的数字，取出对应的str，把str进行回溯。