LeetCode_17 Letter Combinations of a Phone Number-优快云博客

本文链接：https://blog.youkuaiyun.com/sbicode/article/details/50598176

本文介绍了一个典型的回溯问题——电话号码字母组合。给定一个数字字符串，返回所有可能的字母组合。文章提供了一种使用回溯算法的解决方案，并附带了完整的Java代码实现。

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Very typical backtracking problem.Here is the link to the original problem. 点击打开链接

Related question: 22 Generate Parentheses 点击打开链接 and 39 Combination Sum 点击打开链接

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.

1. Backtracking solution:

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<String>();
        if(digits == null || digits.length() == 0) return res;
        String[] mapping = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        char[] cDigits = digits.toCharArray();
        StringBuffer cur = new StringBuffer();
        helper(res, cur, cDigits, 0, mapping);
        return res;
    }
    private void helper(List<String> res, StringBuffer cur, char[] cDigits, int index, String[] mapping){
        if(index == cDigits.length)
            res.add(cur.toString());
        else{
            char cDigit = cDigits[index];
            char[] lowcase = mapping[cDigit-'0'].toCharArray();
            for(char tmp : lowcase){
                cur.append(tmp);
                helper(res, cur, cDigits, index+1, mapping);
                cur.deleteCharAt(cur.length()-1);
            }
        }
    }
}