leetcode-letter combination of a phone number
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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.
public class Solution { //store the mapping String[] map={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; List<String> result=new ArrayList<String>(); public List<String> letterCombinations(String digits) { if(digits.length()==0){ result.add(new String()); return result; }; findCombination(digits,0,new StringBuffer()); return result; } public void findCombination(String digits,int pos, StringBuffer s){ int num=digits.charAt(pos)-'0'; int len=map[num].length(); // length of the string the digit refers to for(int i=0;i<len;i++){ s.append(map[num].charAt(i)); if(pos==digits.length()-1)result.add(s.toString()); else if(pos<digits.length()-1){findCombination(digits,pos+1,s);} /*undo effect*/ s.deleteCharAt(pos); //how to use } }}想法非常简单:用string数组储存电话键盘上面的数字,注意对应的数组下标要和数字对应。
backtracking传递参数,第几个数字,string。
注意的问题:stringbuffer有toString()方法,builder没有。然后delete方法不要忘记。
2在没有输入的情况下传递一个包含空的字符串result。
3charAt ()返回来的是一个char 对应的二进制,不要忘了减去'0'转换成想要的数字。
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