Leetcode: 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"].

像这种DFS的题目，常见的写法无外乎两种，使用recursion, 或者用Stack。本文采用了Recursion的方式。做完后积累的经验有：像这种在一个ArrayList里面罗列可能的path的题目，recursion的参数一般包括：包含最终结果的集合（ArrayList），input（String），递归层次level（int），某一条具体的path（String）。最后这个参数虽然不是必须，但是如果使用了它，将会使recursion非常好写：所有关于这条路径的添加、删除、修改都可以以这个具体的path为操作对象，并且一旦条件满足，就可以把这个path添加到最终的结果集合里面去，用ArrayList add函数

还有一些细节需要注意：用switch函数的时候，要记得写break，否则所有的case都会执行

public class Solution {
public ArrayList<String> letterCombinations(String digits) {
ArrayList<String> res = new ArrayList<String>();
getlist(digits, res, "", 0);
return res;
}

public void getlist(String digits, ArrayList<String> res, String single, int level) {
if (digits.length() == single.length()) {
res.add(single);
return;
}
int c = (int)(digits.charAt(level) - '0'); //get the specific digit at the level unit
String temp = possible(c);
char[] letterpool = temp.toCharArray();
for (char st : letterpool) {
single += st;
getlist(digits, res, single, level+1);
single = single.substring(0, single.length() - 1);
}
}

public String possible(int c) {
String letters = "";
switch (c) {
case 2: letters = "abc"; break;
case 3: letters = "def"; break;
case 4: letters = "ghi"; break;
case 5: letters = "jkl"; break;
case 6: letters = "mno"; break;
case 7: letters = "pqrs"; break;
case 8: letters = "tuv"; break;
case 9: letters = "wxyz"; break;
case 0: letters = " "; break;
}
return letters;
}
}

推荐做法：与上类似但是比上面写的好

public class Solution {
public List<String> letterCombinations(String digits) {
ArrayList<String> res = new ArrayList<String>();
if (digits==null || digits.length()==0) return res;
getlist(digits, res, "", 0);
return res;
}

public void getlist(String digits, ArrayList<String> res, String ss, int level) {
if (level == digits.length()) {
res.add(ss);
return;
}

String cur = interpret(digits.charAt(level));
for (int i=0; i<cur.length(); i++) {
char temp = cur.charAt(i);
getlist(digits, res, ss+temp, level+1);
}
}

public String interpret(char c) {
String result = "";
switch (c) {
case '2': result = "abc"; break;
case '3': result = "def"; break;
case '4': result = "ghi"; break;
case '5': result = "jkl"; break;
case '6': result = "mno"; break;
case '7': result = "pqrs"; break;
case '8': result = "tuv"; break;
case '9': result = "wxyz"; break;
default: result = "";
}
return result;
}
}

另外一种做法，不用DFS和Recursion的想法，用Iteration, 稍微麻烦一点，但是也是一种方法。

public class Solution {
public List<String> letterCombinations(String digits) {
ArrayList<String> res = new ArrayList<String>();
if (digits == null) return res;
if (digits.length() == 0) {
res.add("");
return res;
}
for (int i=0; i<digits.length(); i++) {
char c = digits.charAt(i);
String cr = interpret(c);
ArrayList<String> newres = new ArrayList<String>();
for (int j=0; j<cr.length(); j++) {
char newelem = cr.charAt(j);
if (res.size() == 0) newres.add(Character.toString(newelem));
else {
for (String item : res) {
item = item + newelem;
newres.add(item);
}
}
}
res = newres;
}
return res;
}

public String interpret(char c) {
String result = "";
switch (c) {
case '2': result = "abc"; break;
case '3': result = "def"; break;
case '4': result = "ghi"; break;
case '5': result = "jkl"; break;
case '6': result = "mno"; break;
case '7': result = "pqrs"; break;
case '8': result = "tuv"; break;
case '9': result = "wxyz"; break;
default: result = "";
}
return result;
}
}

注意15行的语法问题，要将一个char类型改成String类型，用到Character.toString(char st), 另外String的‘+’连接是可以连接一个String以及char的