LeetCode_17---Letter Combinations of a Phone Number

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.
翻译：import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

/**
* @author MohnSnow
* @time 2015年6月4日 下午2:10:59
* @title String Backtracking
*/
public class LeetCode17 {

/**
* @param argsmengdx
* -fnst
*/
//第一想法竟然是最Brute的方法，计算字符串的长度然后循环，面壁去。
//数据结构太差了，这个题目需要好好研究一下
//244msA
public static List<String> letterCombinations(String digits) {
List<String> result = new ArrayList<String>();
if (digits.isEmpty()) {
return result;
}
String[] map = new String[] { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
int[] number = new int[digits.length()];//存储digits中每个字符在循环中的编号，类似与进制计算
int k = digits.length() - 1;
while (k >= 0) {
k = digits.length() - 1;
char[] charTemp = new char[digits.length()];//存放临时字符串
for (int i = 0; i < digits.length(); i++) {
charTemp[i] = map[digits.charAt(i) - '0'].charAt(number[i]);//字符串相应位置与临时数组位置对应
}
result.add(new String(charTemp));
while (k >= 0) {//进位
if (number[k] < (map[digits.charAt(k) - '0'].length() - 1)) {//不存在进位
number[k]++;
break;//存在进位才会退出循环
} else {//存在进位
number[k] = 0;
k--;//进位
}
}
}
return result;
}

//https://leetcode.com/discuss/24431/my-java-solution-with-fifo-queue
//244msA---先进先出---仔细研究一下
public static List<String> letterCombinations1(String digits) {
LinkedList<String> ans = new LinkedList<String>();
String[] map = new String[] { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
if (digits.isEmpty())
return ans;
else
ans.add("");
for (int i = 0; i < digits.length(); i++) {
int x = digits.charAt(i) - '0';
while (ans.peek().length() == i) {
String t = ans.remove();
for (char s : map[x].toCharArray())
ans.add(t + s);
}
}
return ans;
}

//http://blog.csdn.net/china_wanglong/article/details/38495355
//深度遍历算法---296msA--递归算法时间复杂度较低
public static List<String> letterCombinations2(String digits) {
List<String> result = new ArrayList<String>();
if (digits.isEmpty())
return result;
String[] map = new String[] { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
char[] middleTemp = new char[digits.length()];
dfsGetStr(digits, 0, middleTemp, map, result);
return result;
}

private static void dfsGetStr(String digits, int index, char[] middleStr, String[] map, List<String> result) {
if (index == digits.length()) {
result.add(new String(middleStr));//middleStr起到一个缓存的作用
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);
}
}

public static void main(String[] args) {
String a = "236";
int x = a.charAt(1) - '0';//字符转换为数字
int y = (int) a.charAt(1);//强制转换字符类型会转换为ASCII码的形式
int z = Character.getNumericValue(a.charAt(1));//作用类似
System.out.println("x:" + x + " y: " + y + " z: " + z);
System.out.println(letterCombinations(a));
System.out.println(letterCombinations1(a));
System.out.println(letterCombinations2(a));
}

}


Code: