LeetCode17 17. Letter Combinations of a Phone Number My Submissions QuestionEditorial Solution

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.

这道题我的思路就是回溯，采用一个堆存当前的字符，当堆的高度和字符串一样长，就是一种输出情况。

public class Solution {

private char[] getLetters(int dig){
if(dig == 0 || dig ==1){
return new char[]{};
}
if(dig == 9){
return new char[]{'w','x','y','z'};
}
if(dig == 7){
return new char[]{'p','q','r','s'};
}
if(dig == 8){
return new char[]{'t','u','v'};
}
char[] result = new char[]{(char)(((dig-2)*3)+'a') , (char)(((dig-2)*3)+'b'), (char)(((dig-2)*3)+'c')};
return result;
}
private List<String> iteraGetLetteCombinations(String digits, Stack<Character> stack, List<String> list){
if(digits.length() == 0){
StringBuilder stringBuilder = new StringBuilder();
for(char c : stack){
stringBuilder.append(c);
}
list.add(stringBuilder.toString());
return list;
}

int dig = digits.charAt(0)-'0';
char[] chars = getLetters(dig);
for(char c: chars){
stack.push(c);
iteraGetLetteCombinations(digits.substring(1,digits.length()), stack, list);
stack.pop();
}
return list;
}

public List<String> letterCombinations(String digits) {
if(digits.length() == 0){
return new ArrayList<String>();
}
Stack<Character> stack = new Stack<>();
List<String> list = new ArrayList<>();
return iteraGetLetteCombinations(digits, stack, list);
}
}为了简化复杂度的计算，我们令每个数字对应3个字符。其中，由数字得到字符，可以采用查表法，更简单，时间复杂度为O（1），而回朔的时间复杂度约为O(3^n)， 空间复杂度为O(3^n); 所以整体的时间复杂度为O（3^n)。

后面发现论坛上有更加简洁的代码如下：

public List<String> letterCombinations(String digits) {
LinkedList<String> ans = new LinkedList<String>();
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;
}时间复杂度和空间复制度和上面的算法一样。