Letter Combinations of a Phone Number--LeetCode

1.题目

Letter Combinations of a Phone Number

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”].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

2.题意

给定一个数字字符串, 返回该数字可以表示的所有可能的字母组合

下面给出了数字到字母的映射 (就像在电话按钮上一样)

string dict[] = {“abc”, “def”, “ghi”, “jkl”, “mno”, “pqrs”, “tuv”, “wxyz”};

注意:

虽然上面的答案是字典的, 但你的答案可能是你想要的任何顺序

3.分析

排列组合问题

1）递归版

建立字典保存每个数字所代表的字符串

用一个变量level，记录当前生成的字符串的字符个数

递归出口为level == digits.size()

注意letterCombinationsDFS(result, dict, digits, level + 1, temp);

不要写成letterCombinationsDFS(result, dict, digits, level + 1, str);

level == digits.size()之后result.push_back(temp);再return;不要写成return result;

2）迭代版

依次读取数字

将数字可以代表的字符加入当前结果

然后进入下一次迭代

注意提前result.push_back(“”);

string temp = result.front();不要写成string temp = result.begin();

不要忘记return reuslt;

4.代码

1）递归版

class Solution {
public:
vector<string> letterCombinations(string digits) {
vector<string> result;

if(digits.size() == 0)
return result;

string dict[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
letterCombinationsDFS(result, dict, digits, 0, "");

return result;
}

void letterCombinationsDFS(vector<string> &result, string dict[],
string digits, int level, string temp)
{
if(level == digits.size())
{
result.push_back(temp);
return;
}

string str = dict[digits[level] - '2'];
for(int i = 0; i < str.size(); ++i)
{
temp.push_back(str[i]);
letterCombinationsDFS(result, dict, digits, level + 1, temp);
temp.pop_back();
}

return;
}
};

2）迭代版

class Solution {
public:
vector<string> letterCombinations(string digits) {
vector<string> result;
if(digits.size() == 0)
return result;

string dict[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
result.push_back("");

for(int i = 0; i < digits.size(); ++i)
{
int n = result.size();
string str = dict[digits[i] - '2'];

for(int j = 0; j < n; ++j)
{
string temp = result.front();
result.erase(result.begin());
for(int k = 0; k < str.size(); ++k)
{
result.push_back(temp + str[k]);
}
}
}

return result;
}
};