Letter Combinations of a Phone Number 电话按键字符的组合@LeetCode

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本文介绍了一种算法，用于根据电话按键上数字与字母的映射关系生成所有可能的字母组合。给定一个数字字符串，通过递归深度优先搜索（DFS）的方式实现，返回所有可能的字母组合。

先搞一张电话数字和字符的对应表，再针对给定电话号码DFS回溯

package Level3;

import java.util.ArrayList;

/**
 * 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.
http://upload.wikimedia.org/wikipedia/commons/thumb/7/73/Telephone-keypad2.svg/200px-Telephone-keypad2.svg.png


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 S17 {

	public static void main(String[] args) {
		System.out.println(letterCombinations("23"));
	}

	public static ArrayList<String> letterCombinations(String digits) {
		String[] ss = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
		
		ArrayList<String> ret = new ArrayList<String>();
		rec(ret, digits.length(), ss, digits, new StringBuffer());
		return ret;
    }
	
	public static void rec(ArrayList<String> ret, int remain, String[] ss, String digits, StringBuffer sb){
		// 说明对digits遍历结束
		if(remain == 0){
			ret.add(sb.toString());
			return;
		}
		
		String s = ss[digits.charAt(0)-'0'];		// 得到digits[0]对应的string s
		// DFS
		for(int i=0; i<s.length(); i++){
			sb = sb.append(s.charAt(i));
			rec(ret, remain-1, ss, digits.substring(1), sb);
			sb.deleteCharAt(sb.length()-1);		// 恢复现场
		}
	}
}

public class Solution {
    
    public List<String> letterCombinations(String digits) {
        String[] map = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        List<String> list = new ArrayList<String>();
        rec(digits, map, list, new StringBuilder(), 0);
        return list;
    }
    
    public void rec(String digits, String[] map, List<String> list, StringBuilder sb, int pos) {
        if(pos == digits.length()) {
            list.add(sb.toString());
            return;
        }
        
        String s = map[digits.charAt(pos)-'0'];
        for(int i=0; i<s.length(); i++) {
            sb.append(s.charAt(i));
            rec(digits, map, list, sb, pos+1);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}