leetcode 17. 电话号码的字母组合

递归

自己写了个递归的算法

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> resList = recursion(digits);
        return resList;
    }

    public List<String> recursion(String str) {
        List<String> resList = new ArrayList<>();
        if (str.isEmpty()) {
            return resList;
        }
        if (str.length() == 1) {
            String s = getStr(str.charAt(0));
            for (int i = 0; i < s.length(); i++) {
                resList.add(s.substring(i, i+1));
            }
            return resList;
        }
        List<String> list = recursion(str.substring(1));
        for (int j = 0; j < list.size(); j++) {
            String s = getStr(str.charAt(0));
            for (int k = 0; k < s.length(); k++) {
                String item = s.charAt(k) + list.get(j);
                resList.add(item);
            }
        }
        return resList;
    }

    public String getStr(char c) {
        String[] buttons = {"", "", "abc",  // 0, 1, 2
                "def", "ghi", "jkl",  // 3, 4, 5
                "mno", "pqrs", "tuv", "wxyz"};  // 6, 7, 8, 9
        int index = c - '0';
        return buttons[index];
    }
}

回溯

题解中给的一种回溯解法。
当题目中出现 “所有组合” 等类似字眼时，我们第一感觉就要想到用回溯。
定义函数 backtrack(combination, nextdigit)，当 nextdigit 非空时，对于 nextdigit[0] 中的每一个字母 letter，执行回溯 backtrack(combination + letter,nextdigit[1:]，直至 nextdigit 为空。最后将 combination 加入到结果中。

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits: return []

        phone = {'2':['a','b','c'],
                 '3':['d','e','f'],
                 '4':['g','h','i'],
                 '5':['j','k','l'],
                 '6':['m','n','o'],
                 '7':['p','q','r','s'],
                 '8':['t','u','v'],
                 '9':['w','x','y','z']}
                
        def backtrack(conbination,nextdigit):
            if len(nextdigit) == 0:
                res.append(conbination)
            else:
                for letter in phone[nextdigit[0]]:
                    backtrack(conbination + letter,nextdigit[1:])

        res = []
        backtrack('',digits)
        return res

java解法

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> combinations = new ArrayList<String>();
        if (digits.length() == 0) {
            return combinations;
        }
        Map<Character, String> phoneMap = new HashMap<Character, String>() {{
            put('2', "abc");
            put('3', "def");
            put('4', "ghi");
            put('5', "jkl");
            put('6', "mno");
            put('7', "pqrs");
            put('8', "tuv");
            put('9', "wxyz");
        }};
        backtrack(combinations, phoneMap, digits, 0, new StringBuffer());
        return combinations;
    }

    public void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index, StringBuffer combination) {
        if (index == digits.length()) {
            combinations.add(combination.toString());
        } else {
            char digit = digits.charAt(index);
            String letters = phoneMap.get(digit);
            int lettersCount = letters.length();
            for (int i = 0; i < lettersCount; i++) {
                combination.append(letters.charAt(i));
                backtrack(combinations, phoneMap, digits, index + 1, combination);
                combination.deleteCharAt(index);
            }
        }
    }
}