017 电话号码的之母组合

LeetCode 第十七题 电话号码的之母组合

给定一个数字字符串，返回数字所有可能表示的字母组合。
下面给出数字到字母的映射（和电话号码一样）。
输入：数字字符串 “23”
输出：[“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

在这里使用了队列思想。对于每一个数字，将其所对应的字母，添加到队列中的每一个元素中。这里参考了解题步骤中的Java代码。

C++

class Solution
{
public:
    vector<string> letterCombinations(string digits)
    {
        vector<string> result=vector<string>();
        if(digits.size()==0)
            return result;
        map<int,string>mapping;
        ///顺便回顾一下map的三种插入方法
        mapping.insert(pair<int,string>(0,"0"));
        mapping.insert(pair<int,string>(1,"1"));
        mapping.insert(pair<int,string>(2,"abc"));
        mapping.insert(map<int,string>::value_type(3,"def"));
        mapping.insert(map<int,string>::value_type(4,"ghi"));
        mapping.insert(map<int,string>::value_type(5,"jkl"));
        mapping[6]="mno";
        mapping[7]="pqrs";
        mapping[8]="tuv";
        mapping[9]="wxyz";
        result.push_back("");
        for(int i=0; i<digits.size(); i++)
        {
            int x = (digits[i]-'0');
            int resultLength = result.size();
            for(int j=0;j<resultLength;j++)
            {
                string s=result[0];
                result.erase(result.begin());
                for(int k=0;k<mapping[x].size();k++)
                {
                    string current_s = s+mapping[x][k];
                    result.push_back(current_s);
                }
            }
        }
        return result;
    }
};

Python

class Solution(object):
    def letterCombinations(self, digits):
        result = []
        if len(digits) == 0:
            return result
        mapping = {0: "0", 1: "1", 2: "abc", 3: "def", 4: "ghi", 5: "jkl", 6: "mno", 7: "pqrs", 8: "tuv", 9: "wxyz"}
        result = [""]
        for i in range(len(digits)):
            x = int(digits[i])
            resultLength = len(result)
            for j in range(resultLength):
                s = result[0]
                result.remove(s)
                for k in range(len(mapping[x])):
                    current_s = s+mapping[x][k]
                    result.append(current_s)
        return result

Java

    public static List<String> letterCombinations(String digits) {
        LinkedList<String> ans = new LinkedList<String>();
        if (digits.isEmpty())
            return ans;
        String[] mapping = new String[] { "0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
        ans.add("");
        for (int i = 0; i < digits.length(); i++) {
            int x = Character.getNumericValue(digits.charAt(i));
            while (ans.peek().length() == i) {
                String t = ans.remove();
                for (char s : mapping[x].toCharArray())
                    ans.add(t + s);
            }
        }
        return ans;
    }