[LeetCode] 017. Letter Combinations of a Phone Number (Mediu
发布时间：2020-12-13 20:16:59 所属栏目：PHP教程 来源：网络整理
导读：索引：[LeetCode] Leetcode 题解索引 (C/Java/Python/Sql) Github: https://github.com/illuz/leetcode 017.Letter_Combinations_of_a_Phone_Number (Medium) 链接 ： 题目：https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/ 代码(
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索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) 017.Letter_Combinations_of_a_Phone_Number (Medium)链接：题目：https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/ 题意：在手机上按字母，给出按的数字键，问所有的按的字母的情况。 分析：DFS 过去是比较轻松的写法。 代码：C++: class Solution {
private:
const string alpha[10] = {
" ","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"
};
void dfs(vector<string> &res,string &ab,string &digits,int cur) {
if (cur >= digits.length()) {
res.push_back(ab);
return;
}
for (auto &a : alpha[digits[cur] - '0']) {
ab.push_back(a);
dfs(res,ab,digits,cur + 1);
ab.pop_back();
}
}
public:
vector<string> letterCombinations(string digits) {
vector<string> res;
string alphas;
dfs(res,alphas,0);
return res;
}
};
Java: public class Solution {
private String[] alpha = new String[] {
" ","wxyz"
};
private StringBuilder word;
private void dfs(List<String> res,String digits,int cur) {
if (cur >= digits.length()) {
res.add(word.toString());
} else {
for (int i = 0; i < alpha[digits.charAt(cur) - '0'].length(); ++i) {
word.append(alpha[digits.charAt(cur) - '0'].charAt(i));
dfs(res,cur + 1);
word.deleteCharAt(word.length() - 1);
}
}
}
public List<String> letterCombinations(String digits) {
List<String> ret = new ArrayList<String>();
word = new StringBuilder();
dfs(ret,0);
return ret;
}
} Python: class Solution:
# @return a list of strings,[s1,s2]
def letterCombinations(self,digits):
alpha = [" ","wxyz"]
res = []
word = []
def dfs(cur):
if cur >= len(digits):
res.append(''.join(word))
else:
for x in alpha[(int)(digits[cur]) - (int)('0')]:
word.append(x)
dfs(cur + 1)
word.pop()
dfs(0)
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