93 lines
2.6 KiB
Java
93 lines
2.6 KiB
Java
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//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。答案可以按 任意顺序 返回。
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//
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// 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。
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// 示例 1：
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//
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//输入：digits = "23"
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//输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
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// 示例 2：
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//输入：digits = ""
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//输出：[]
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// 示例 3：
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//
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//
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//输入：digits = "2"
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//输出：["a","b","c"]
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// 提示：
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//
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// 0 <= digits.length <= 4
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// digits[i] 是范围 ['2', '9'] 的一个数字。
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//
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// Related Topics 哈希表 字符串 回溯 👍 1453 👎 0
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package leetcode.editor.cn;
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import java.util.ArrayList;
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import java.util.HashMap;
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import java.util.List;
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import java.util.Map;
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//17:电话号码的字母组合
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class LetterCombinationsOfAPhoneNumber {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new LetterCombinationsOfAPhoneNumber().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public List<String> letterCombinations(String digits) {
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List<String> combinations = new ArrayList<>();
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if (digits.length() == 0) {
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return combinations;
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}
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Map<Character, String> phoneMap = new HashMap<Character, String>() {{
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put('2', "abc");
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put('3', "def");
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put('4', "ghi");
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put('5', "jkl");
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put('6', "mno");
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put('7', "pqrs");
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put('8', "tuv");
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put('9', "wxyz");
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}};
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backtrack(combinations, phoneMap, digits, 0, new StringBuffer());
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return combinations;
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}
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public void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index, StringBuffer combination) {
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if (index == digits.length()) {
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combinations.add(combination.toString());
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} else {
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char digit = digits.charAt(index);
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String letters = phoneMap.get(digit);
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int lettersCount = letters.length();
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for (int i = 0; i < lettersCount; i++) {
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combination.append(letters.charAt(i));
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backtrack(combinations, phoneMap, digits, index + 1, combination);
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combination.deleteCharAt(index);
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}
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}
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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