Languages: Java
Categories: 算法分析
gistfile1.txt
//给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
//
// 给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
//
//
//
// 示例:
//
// 输入："23"
//输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
//
//
// 说明:
//尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。
// Related Topics 字符串 回溯算法

import java.util.ArrayList;
import java.util.List;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
private static final char[][] sMap = new char[10][];

static {
sMap[2] = new char[]{'a', 'b', 'c'};
sMap[3] = new char[]{'d', 'e', 'f'};
sMap[4] = new char[]{'g', 'h', 'i'};
sMap[5] = new char[]{'j', 'k', 'l'};
sMap[6] = new char[]{'m', 'n', 'o'};
sMap[7] = new char[]{'p', 'q', 'r', 's'};
sMap[8] = new char[]{'t', 'u', 'v'};
sMap[9] = new char[]{'w', 'x', 'y', 'z'};
}

private char[] mChars;
private List<String> mResult;
private int mTarget;
private String mDigits;

public List<String> letterCombinations(String digits) {
if (digits == null || digits.isEmpty()) {
return new ArrayList<>();
}

mDigits = digits;
mChars = new char[digits.length()];
mTarget = digits.length();
mResult = new ArrayList<>();

findSolution(0);

return mResult;
}

private void findSolution(int index) {
if (index == mTarget) {
return;
}

char[] chars = sMap[charToNum(mDigits.charAt(index))];

for (int j = 0; j < chars.length; j++) {
mChars[index] = chars[j];
findSolution(index + 1);
}
}

private int charToNum(char c) {
return c - '0';
}
}
//leetcode submit region end(Prohibit modification and deletion)