leetcode:17. Letter Combinations of a Phone Number

描述

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string “23”
Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

思路

思路比较简单，就是不断做笛卡尔积。

代码

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> combineVec;        if(digits.empty()) return combineVec;         static const string letterMap[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        combineVec.push_back("");        int index;        for(int i = 0; i < digits.size(); i++){            index = digits[i] - '0';            if(index < 2 || index > 9) continue;            vector<string> temp;             const string& letterMapValue = letterMap[index];            for(int j = 0; j < combineVec.size();j++){                for(int k = 0; k < letterMapValue.size(); k++){                    temp.push_back(combineVec[j] + letterMapValue[k]);                }            }            combineVec=temp;        }        return combineVec;    }};

结果

他山之玉

C++ O(n) solutioin

vector<string> letterCombinations(string digits) {    vector<string> result;    if(digits.empty()) return vector<string>();    static const vector<string> v = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    result.push_back("");   // add a seed for the initial case    for(int i = 0 ; i < digits.size(); ++i) {        int num = digits[i]-'0';        if(num < 0 || num > 9) break;        const string& candidate = v[num];        if(candidate.empty()) continue;        vector<string> tmp;        for(int j = 0 ; j < candidate.size() ; ++j) {            for(int k = 0 ; k < result.size() ; ++k) {                tmp.push_back(result[k] + candidate[j]);            }        }        result.swap(tmp);    }    return result;}

在第二句的时候，我用的是digits.size()来判断是否为空，结果时间差了2ms。这个地方下次要注意了。

Java O(n) solution

public class Solution {    public List<String> letterCombinations(String digits) {        List<String> list = new ArrayList<>();        if (digits == null || digits.trim().length() == 0) return list;        String[] str = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};        char[] ch = digits.trim().toCharArray();        int sum = 1;        String [] arr = new String[ch.length];        for (int i = 0; i < ch.length; i++) {           arr[i] = str[ch[i] - '0'];        }        list = new ArrayList<>(sum);       return  combination(arr, list, 0, "");    }    private List<String> combination(String[] s, List<String> list, int index, String result) {        if (result.length() == s.length) {            list.add(result);            return list;        }        for (int i = index; i < s.length; i++) {            for (int j = 0; j < s[i].length(); j++) {               combination(s,list, i + 1, result + s[i].substring(j, j + 1));            }        }        return list;     }

这个是利用递归来做的笛卡尔积。

python O(n) solution

class Solution:    # @return a list of strings, [s1, s2]    def letterCombinations(self, digits):        # define a dictionary of the character <-> number projection        pad = {'2':'abc', '3':'def', '4':'ghi', '5':'jkl', '6':'mno', '7':'pqrs', '8':'tuv', '9':'wxyz'}        # if the input is empty        if len(digits) == 0:            return [""]        # for number string length equals n-1            d0 = digits[:-1]        comb0 = self.letterCombinations(d0)        # combine the last digit with previous n-1 string        d1 = digits[-1]        string1 = pad[d1]        comb1 = []        #  consider every combinations        for s in string1:            for c in comb0:                cs = c+s                comb1.append(cs)        return comb1

反思

先写出来，再优化