【leetcode】Letter Combinations of a Phone Number

• 建一个二位char数组对应2,3,4……9的char值，注意2对应abc，9对应wxyz
• 对于输入的string用charAt取出对应数，注意要减去‘0‘才是int数，因为是错位对应，从2开始，所以还要减去1，因为数组索引，所以还要减去1
• 需要用到递归
• 如果取数字的位置i大于等于电话号码的长度，将sb中的数tostring添加到返回ret中
• 删掉最后一个字母，换另外一个字母
• 如果最后一个字母遍历完，遍历倒数第二个字母
• 最终遍历完成输出ret
• ret是链表

  package leetcode;

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  ? ?

  import java.util.ArrayList;

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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"].

  */

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  public class LetterCombinationsofaPhoneNumber {

  private char[][] map = new char[][] { { ‘a‘, ‘b‘, ‘c‘ }, { ‘d‘, ‘e‘, ‘f‘ },

  { ‘g‘, ‘h‘, ‘i‘ }, { ‘j‘, ‘k‘, ‘l‘ }, { ‘m‘, ‘n‘, ‘o‘ },

  { ‘p‘, ‘q‘, ‘r‘, ‘s‘ }, { ‘t‘, ‘u‘, ‘v‘ }, { ‘w‘, ‘x‘, ‘y‘, ‘z‘ } };

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  public ArrayList<String> letterCombinations(String digits) {

  ArrayList<String> ret = new ArrayList<String>();

  StringBuilder sb = new StringBuilder();

  letterCombinations(digits, 0, sb, ret);

  return ret;

  }

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  private void letterCombinations(String digits, int i, StringBuilder sb,

  ArrayList<String> ret) {

  if (i >= digits.length()) {

  ret.add(sb.toString());

  } else {

  int index = digits.charAt(i) - ‘1‘ - 1;

  int size = map[index].length;

  for (int j = 0; j < size; j++) {

  sb.append(map[index][j]);

  letterCombinations(digits, i + 1, sb, ret);

  sb.deleteCharAt(sb.length() - 1);

  }

  }

  }

  }

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【leetcode】Letter Combinations of a Phone Number

原文：http://www.cnblogs.com/keedor/p/4366747.html