2019.1.28
Given a string containing digits from 2-9
inclusive, 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. Note that 1 does not map to any letters.
Example:
Input: "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.
这道题是按照老式手机九键键盘的数字与字母的对应规则,求出一组全是数字的字符串的对应的字母的全部排列组合。
解法一:递归+字典
我们首先建立一个字典,用来存放数字与字符串的对应关系,在递归过程中,我们设置一个depth记录当前已经扫描到第几个数字了,若depth已经是给定的个数了,那就将当前的组合加入结果res中,再直接return,若不是,则将该数字对应的字符串从字典中取出来,再循环遍历该字符串,将字符串中的字符依次添加到当组合的后面,再递归。
当然,LeetCode的测试用例基本都有空,注意一下特殊情况即可。
C++代码:
class Solution {
public:
vector<string> letterCombinations(string digits) {
if (digits.size()==0) return {};
vector<string> res;
string dict[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
letterCombinationsDFS(digits, dict, 0, "", res);
return res;
}
void letterCombinationsDFS(string digits, string dict[], int depth, string out, vector<string> &res) {
if (depth == digits.size()) {res.push_back(out); return;}
string str = dict[digits[depth] - '0'];
for (int i = 0; i < str.size(); ++i) {
letterCombinationsDFS(digits, dict, depth + 1, out + str[i], res);
}
}
};
解法二:迭代
我们建立一个临时的字符串数组temp,在遍历digits时,每次从dict中取出字符串str,然后遍历取出字符串中的所有字符,再遍历当前结果res中的每一个字符串,将字符加到后面,并加入到临时字符串数组temp中。取出的字符串str遍历完成后,将临时字符串数组赋值给结果res
C++代码:
class Solution {
public:
vector<string> letterCombinations(string digits) {
if (digits.empty()) return {};
vector<string> res{""};
string dict[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
for (int i = 0; i < digits.size(); ++i) {
vector<string> temp;
string str = dict[digits[i] - '0'];
for (int j = 0; j < str.size(); ++j) {
for (string s : res) temp.push_back(s + str[j]);
}
res = temp;
}
return res;
}
};