Lc17-电话号码的字母组合

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

/*
 * Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

Example:

Input: "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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */
public class Lc17 {
    public static List<String> letterCombinations(String digits) {
        // 避免频繁扩容
        LinkedList<String> ans = new LinkedList<String>();
        String[] mapping = new String[] { "0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

        // 空判断
        if (digits.length() == 0) {
            return new ArrayList<>();
        }
        // 这里是利用队列的特性进行bfs遍历
        ans.add("");
        // 由于不知道具体由多少个数字，所以用for控制
        for (int i = 0; i < digits.length(); i++) {
            // 找到对应数字 ，比如输入23 ，找到2或者3
            int x = Character.getNumericValue(digits.charAt(i));
            /*
             * 当第0次循环时，队列中的元素长度时空 即0位
             * 当第一次循环时，队列中元素长度时a/b/c，即1位。以此类推
             */
            while (ans.peek().length() == i) {
                // 获取队列中的元素，并移除该元素重新组合
                String t = ans.remove();
                for (Character ch : mapping[x].toCharArray()) {
                    ans.add(t + ch);
                }
            }
        }
        return ans;
    }

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

原文地址：https://www.cnblogs.com/xiaoshahai/p/12182608.html