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给一个数字字符串，按照电话上各个数字对应的字母，返回该字符串代表的所有可能的字母组合。
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"].
考虑特殊情况：
依次遍历字符串中的每个数，寻找可能的组合。
public List<String> letterCombinations( String digits ){ String[] map = { " ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; int len = digits.length(); LinkedList<String> res = new LinkedList<String>(); if( len<1 ) return res; res.add(""); for( int i=0; i<len; i++ ){ int now = digits.charAt( i ) - ‘0‘; if( now<2 ) continue; char[] toInsert = map[now].toCharArray(); while( res.peek().length() == i ) { String each = res.remove(); for( int j=0; j<toInsert.length; j++ ) res.add( each + toInsert[j] ); } } return res; }
对数字进行解析，相当于遍历一棵树，可以使用数的遍历算法
public List<String> letterCombinations(String digits) { List<String> result = new ArrayList<String>(); 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); } }
Letter Combinations of a Phone Number
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原文地址：http://www.cnblogs.com/hf-cherish/p/4607097.html
