17. Letter Combinations of a Phone Number
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"].
public class Solution { public List<String> letterCombinations(String digits) { //存放答案的list List<String> list=new ArrayList<String>(); //建立map存放手机按键 Map<String, String> map=new HashMap<String, String>(); map.put("2", "abc"); map.put("3", "def"); map.put("4", "ghi"); map.put("5", "jkl"); map.put("6", "mno"); map.put("7", "pqrs"); map.put("8", "tuv"); map.put("9", "wxyz"); //循环判断 for (int i = 0; i < digits.length(); i++) { String s=digits.substring(i,i+1); //如果能查到数字键盘 //if (map.get(s)!=null) { //如果list不为空,即并非刚开始 if (list.size()!=0) { //记住目前list的长度 int size=list.size(); //循环list当前长度 for (int j = 0; j < size; j++) { //将list中的每个值与接下来的对应数字键盘的可能性进行拼接 for (int k = 0; k <map.get(s).length(); k++) { String a=list.get(j)+map.get(s).charAt(k); list.add(a); } } //删掉之前的list for (int j = size-1; j >=0; j--) { list.remove(j); } }else { for (int j = 0; j < map.get(s).length(); j++) { String a=map.get(s).charAt(j)+""; list.add(a); } } //} } return list; } }
https://www.youtube.com/watch?v=uMmFXWs_ZMY
public class Solution { public List<String> letterCombinations(String digits) { List<String> result=new ArrayList<String>(); if(digits==null||digits.length()==0||digits.contains("0")||digits.contains("1")){ return result; } String[] map=new String[10]; map[0]=""; map[1]=""; map[2]="abc"; map[3]="def"; map[4]="ghi"; map[5]="jkl"; map[6]="mno"; map[7]="qprs"; map[8]="tuv"; map[9]="wxyz"; StringBuilder sb=new StringBuilder(); backtrack(sb,digits,map,result,0); return result; } public void backtrack(StringBuilder sb,String digits,String[] map,List<String> result,int start){ if(sb.length()==digits.length()){ result.add(sb.toString()); return; } for(int i=start;i<digits.length();i++){ int tt=digits.charAt(i) - '0'; for(int j=0;j<map[tt].length();j++){ sb.append(map[tt].charAt(j)); backtrack(sb,digits,map,result,i+1); sb.deleteCharAt(sb.length()-1);//剪枝 } } } }
回溯法,精华是剪枝