leetcode -- Letter Combinations of a Phone Number -- 重点

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

先手工枚举，构造解空间树.标准dfs求解。

class Solution(object):    def dfs(self, digits, target_lvl, mydict, subres, res):        lvl = len(subres)        if lvl == target_lvl:            if subres:                res.append(subres[:])            return        else:            for x in mydict[digits[lvl]]:#子节点为下一个数字对应的每个字母                self.dfs(digits, target_lvl, mydict, subres + x, res)    def letterCombinations(self, digits):        """        :type digits: str        :rtype: List[str]        """        mydict = {'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']        }        res = []        self.dfs(digits, len(digits), mydict, "",res)        return res

自己重写code

这里的子集树，每一层对应一个数字的三个字母，其子节点为下一个数字的三个字母。

class Solution(object):    def dfs(self, candidates, level, mydict, subres, res):        if level == len(candidates):            res.append(subres)            return#不要忘记return        for j in mydict[candidates[level]]:#每一个level对应一个数字            self.dfs(candidates, level + 1, mydict, subres + j, res)    def letterCombinations(self, digits):        """        :type digits: str        :rtype: List[str]        """        mydict = {'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']        }        if not digits: return []        res = []        self.dfs(digits, 0, mydict, '', res)        return res