Medium 17题 Letter Combinations of a Phone Number

最新推荐文章于 2022-04-25 11:56:55 发布

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本文介绍了一种通过递归实现的算法，用于根据电话按键上对应的字母，找出所有可能的字母组合。输入为一个数字字符串，输出是该数字对应的所有可能的字母组合列表。

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Question:

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.

Solution:

public class Solution {
    private static final String[] KEYS = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
    public List<String> letterCombinations(String digits) {
        List<String> ret = new LinkedList<String>();
    	combination("", digits, 0, ret);
    	return ret;
    }
    public void combination(String prefix, String digits, int offset, List<String> ret){
        if(offset==digits.length())
        {
            ret.add(prefix);
            return;
        }
        String letter=KEYS[digits.charAt(offset)-'0'];
        for(int i=0;i<=letter.length()-1;i++)
        {
            combination(prefix+letter.charAt(i),digits,offset+1,ret);
        }
    }
}

需要加一个判断：

public class Solution {
    private static final String[] KEYS = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
    public List<String> letterCombinations(String digits) {
        List<String> ans=new LinkedList<String>();
        if(digits.length()==0)
            return ans;
        combination("",digits,0,ans);
        return ans;
    }
    public void combination(String prefix, String digits, int offset, List<String> ans){
        if(digits.length()==offset)
        {
            ans.add(prefix);
            return;
        }
        String p=KEYS[digits.charAt(offset)-'0'];
        for(int i=0;i<=p.length()-1;i++)
        {
            combination(prefix+p.charAt(i),digits,offset+1,ans);
        }
    }
}

11/19

comb(res,digits,tmp+letter.charAt(i),offset+1); //++offset,和offset+1差别

以及另一种解法

 LinkedList<String> res = new LinkedList<String>();
        if(digits.length()==0)
            return res;
        res.add("");
        String[] KEY = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        for(int i =0; i<=digits.length()-1;i++){
            String letter = KEY[digits.charAt(i)-'0'];
            while(res.peek().length()==i){
                String tmp = res.remove();
                for(int j=0;j<=letter.length()-1;j++)
                    res.add(tmp+letter.charAt(j));
            }
        }
        return res;