代码随想录算法训练营第24, 25天

class Solution {
    List<List<Integer>> ans = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> combine(int n, int k) {
        backtrace(1, n, k);
        return ans;
    }
    // [start, end]
    public void backtrace(int start, int end, int k) {
        if(path.size() == k) ans.add(new ArrayList<>(path));
        else {
            for(int i = start; i <= end - (k - path.size()) + 1; i++) {
                System.out.println("=");
                path.add(i);
                backtrace(i + 1, end, k);
                path.removeLast();
            }
        }
    }
}

class Solution {
    public LinkedList<Integer> path = new LinkedList<>();
    public int now_sum = 0;
    public List<List<Integer>> ans = new ArrayList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
        backtrack(k, n, 1, 9);
        return ans;
    }
    public void backtrack(int k, int target, int start, int end) {
        if(path.size() == k) {
            if(now_sum == target) ans.add(new ArrayList(path));
        }
        else {
            for(int i = start; i <= end - (k - path.size()) + 1 && i <= (target - now_sum); i++) {
                path.add(i);
                now_sum += i;
                backtrack(k, target, i + 1, end);
                path.removeLast();
                now_sum -= i;
            }
        }
    }
}

class Solution {
    List<List<Character>> table;
    List<String> ans = new ArrayList<>();
    LinkedList<Character> path = new LinkedList<>();
    public List<String> letterCombinations(String digits) {
        if (digits == null || digits.length() == 0) return ans;
        init_table();
        backtrack(digits, 0);
        return ans;
    }
    public void backtrack(String digits, int digit_index) {
        if(path.size() == digits.length()) ans.add(path.stream().map(String::valueOf).collect(Collectors.joining()));
        else {
            List<Character> chars4num = table.get(digits.charAt(digit_index) - '0');
            for(int i = 0; i < chars4num.size(); i++) {
                path.add(chars4num.get(i));
                backtrack(digits, digit_index + 1);
                path.removeLast();
            }
        }
    }
 
    public void init_table() {
        table = Arrays.asList(
            Arrays.asList(),
            Arrays.asList(),
            Arrays.asList('a', 'b', 'c'),
            Arrays.asList('d', 'e', 'f'),
            Arrays.asList('g', 'h', 'i'),
            Arrays.asList('j', 'k', 'l'),
            Arrays.asList('m', 'n', 'o'),
            Arrays.asList('p', 'q', 'r', 's'),
            Arrays.asList('t', 'u', 'v'),
            Arrays.asList('w', 'x', 'y', 'z')
        );
    }
}

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;
        }
        //初始对应所有的数字，为了直接对应2-9，新增了两个无效的字符串""
        String[] numString = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        //迭代处理
        backTracking(digits, numString, 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 temp = new StringBuilder();
 
    //比如digits如果为"23",num 为0，则str表示2对应的 abc
    public void backTracking(String digits, String[] numString, int num) {
        //遍历全部一次记录一次得到的字符串
        if (num == digits.length()) {
            list.add(temp.toString());
            return;
        }
        //str 表示当前num对应的字符串
        String str = numString[digits.charAt(num) - '0'];
        for (int i = 0; i < str.length(); i++) {
            temp.append(str.charAt(i));
            //c
            backTracking(digits, numString, num + 1);
            //剔除末尾的继续尝试
            temp.deleteCharAt(temp.length() - 1);
        }
    }
}