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问题描述
给定一个数字串,返回这个数组串所表示的所有可能的字母组合。
数字和字母的映射关系类似于一个手机的拨号键盘,如图所示:例如:输入数字串“23”
返回: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].注意:虽然上述答案在返回时是按照字典字母顺序排列的,但是你可以按照你想要的顺序返回结果。
算法设计
代码片段(一种求解算法)
public static ListletterCombinations(String digits) { List result =new ArrayList<>(); if(digits == null || digits.isEmpty()) return result; result.add(""); String []btns = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; for(int i =0 ; i < digits.length() ;i++) { List res = new ArrayList<>(); String letter = btns[digits.charAt(i)-'0']; for(int j = 0 ; j < result.size();j++)//遍历上一个列表,取出每一个元素,并和新的元素的每一个字符加起来保存 { for(int k = 0; k< letter.length(); k++)//遍历当前数字对应的所有字符 { res.add(result.get(j)+letter.charAt(k)); } } result = res; } return result; }
下面再给出不同的求解算法
package com.bean.algorithmbasic;import java.util.ArrayList;import java.util.Iterator;import java.util.List;public class LetterCombinationsDemo2 { public static ListletterCombinations(String digits) { String digitletter[] = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" }; List result = new ArrayList (); if (digits.length() == 0) return result; result.add(""); for (int i = 0; i < digits.length(); i++) result = combine(digitletter[digits.charAt(i) - '0'], result); return result; } public static List combine(String digit, List l) { List result = new ArrayList (); for (int i = 0; i < digit.length(); i++) for (String x : l) result.add(x + digit.charAt(i)); return result; } public static void main(String[] args) { // TODO Auto-generated method stub LetterCombinationsDemo2 lcd = new LetterCombinationsDemo2(); List list = lcd.letterCombinations("23"); Iterator itx = list.iterator(); while (itx.hasNext()) { String str = itx.next(); System.out.println(str); } }}
解法二:
package com.bean.algorithmbasic;import java.util.ArrayList;import java.util.Iterator;import java.util.List;public class LetterCombinationsDemo4 { /* * 算法设计分析 * 数字0-9对应的ASCII码值:依次为:48,49,50,51,52,53,54,55,56,57 * */ public ListletterCombinations(String digits) { List list = new ArrayList<>(); // if (digits == null || digits.trim().length() == 0) return list; String[] str = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; //将字符串digits转化为字符数组,例如给定“23”,将转化为字符数组'2','3' char[] ch = digits.trim().toCharArray(); int sum = 1; String [] arr = new String[ch.length]; for (int i = 0; i < ch.length; i++) { //从数组str中取出对应的字符串 // 例如 '2'-'0',得到结果 2,那么 str[2]则对应于"abc"; '3'-'0',得到结果3,那么str[3]则对应于"def" arr[i] = str[ch[i] - '0']; System.out.print(arr[i]+"\t"); } System.out.println(); list = new ArrayList<>(sum); //调用组合方法,并返回处理结果 return combination(arr, list, 0, ""); } /* * combination方法的作用 * 输入参数: * String[] s * List list * int index * String result * * */ private List combination(String[] s, List list, int index, String result) { if (result.length() == s.length) { list.add(result); return list; } for (int i = index; i < s.length; i++) { for (int j = 0; j < s[i].length(); j++) { //递归调用,做笛卡尔积 combination(s,list, i + 1, result + s[i].substring(j, j + 1)); } } return list; } public static void main(String[] args) { // TODO Auto-generated method stub LetterCombinationsDemo4 lcd = new LetterCombinationsDemo4(); List list = lcd.letterCombinations("23"); Iterator itx = list.iterator(); while (itx.hasNext()) { String str = itx.next(); System.out.println(str); } }}
(完)