LeetCode-17 Letter Combinations of a Phone Number(手机拨码-DFS)

LeetCode-17 Letter Combinations of a Phone Number(手机拨码-DFS)

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.

public class Solution {    List<String> list = new ArrayList<String>();    public List<String> letterCombinations(String digits) {if(digits.equals("")) {return list;}    String[] map = new String[10];         map[0] = "";          map[1] = "";          map[2] = "abc";          map[3] = "def";          map[4] = "ghi";          map[5] = "jkl";          map[6] = "mno";          map[7] = "pqrs";          map[8] = "tuv";          map[9] = "wxyz";         String[] mstr = new String[digits.length()];        for (int i = 0; i < digits.length(); i++) {        mstr[i] = map[digits.charAt(i)-48];}        print(mstr, 0, mstr.length, "");        return list;    }    public void print(String mstr[],int l,int n,String str){if (n <= 0) {list.add(str);} else {for (int i = 0; i < mstr[l].length(); i++) {print(mstr, l+1, n-1, str + mstr[l].charAt(i) + "");}}    }}

Runtime: 228 ms

深度优先搜索，以上是递归的写法。

看到别人的迭代的写法，在此也记录下来。

public class Solution {//DFS 迭代的方法，记得研究一下。public List<String> letterCombinations(String digits) {List<String> result = new ArrayList<String>();if(digits.equals("")) {return result;}String[] map = new String[10];map[0] = "";map[1] = "";map[2] = "abc";map[3] = "def";map[4] = "ghi";map[5] = "jkl";map[6] = "mno";map[7] = "pqrs";map[8] = "tuv";map[9] = "wxyz";int[] number = new int[digits.length()];//存储digits中每个字符在循环中的编号，初始为0int k = digits.length()-1;while(k>=0) {k = digits.length()-1;char[] charTemp = new char[digits.length()];for(int i=0; i<digits.length(); i++) {charTemp[i] = map[digits.charAt(i)-'0'].charAt(number[i]);}result.add(new String(charTemp));while(k>=0) {if( number[k] < (map[digits.charAt(k)-'0'].length()-1) ) {number[k]++;break;} else {number[k] = 0;k--;}}}return result;    }}