LeetCode----17. Letter Combinations of a Phone Number（遍历树+多重循环）

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.

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遍历树：for  +  递归

Input:""

Output:[""]

Expected:[]

public class Solution {public List<String> letterCombinations(String digits) {                  List<String> result = new ArrayList<String>();                  if(digits==null||digits.length()==0){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] = "pqrs";          map[8] = "tuv";          map[9] = "wxyz";          char[]  middleTemp = new char[digits.length()];          dfsGetStr(digits, 0, middleTemp, map, result);           return result;      }            private void dfsGetStr(String digits,int index,               char[] middleStr, String[] map, List<String> result) {          if(index == digits.length()) {              result.add(new String(middleStr));              return ;          }          char strChar = digits.charAt(index);          for(int i=0; i<map[strChar-'0'].length(); i++) {              middleStr[index] = map[strChar-'0'].charAt(i);              dfsGetStr(digits, index+1, middleStr, map, result);          }      }  }

普通循环：

public List<String> letterCombinations(String digits) {List<String> result = new ArrayList<String>();if(digits.isEmpty()) {result.add("");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] = "pqrs";map[8] = "tuv";map[9] = "wxyz";int[] number = new int[digits.length()];//存储digits中每个字符在循环中的编号，初始为0int k = digits.length()-1;while(k>=0) {k = digits.length()-1;char[] charTemp = new char[digits.length()];for(int i=0; i<digits.length(); i++) {charTemp[i] = map[digits.charAt(i)-'0'].charAt(number[i]);}result.add(new String(charTemp));while(k>=0) {if( number[k] < (map[digits.charAt(k)-'0'].length()-1) ) {number[k]++;break;} else {number[k] = 0;k--;}}}return result;    }

public class Solution {    public ArrayList<String> letterCombinations(String digits) {        ArrayList<String> result = new ArrayList<String>();        if (digits == null || digits.equals("")) {            return result;        }        Map<Character, char[]> map = new HashMap<Character, char[]>();        map.put('0', new char[] {});        map.put('1', new char[] {});        map.put('2', new char[] { 'a', 'b', 'c' });        map.put('3', new char[] { 'd', 'e', 'f' });        map.put('4', new char[] { 'g', 'h', 'i' });        map.put('5', new char[] { 'j', 'k', 'l' });        map.put('6', new char[] { 'm', 'n', 'o' });        map.put('7', new char[] { 'p', 'q', 'r', 's' });        map.put('8', new char[] { 't', 'u', 'v'});        map.put('9', new char[] { 'w', 'x', 'y', 'z' });        StringBuilder sb = new StringBuilder();        helper(map, digits, sb, result);        return result;    }    private void helper(Map<Character, char[]> map, String digits,         StringBuilder sb, ArrayList<String> result) {        if (sb.length() == digits.length()) {            result.add(sb.toString());            return;        }        for (char c : map.get(digits.charAt(sb.length()))) {            sb.append(c);            helper(map, digits, sb, result);            sb.deleteCharAt(sb.length() - 1);        }    }}