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LeetCode-500-键盘行

 2 years ago
source link: https://segmentfault.com/a/1190000040775038
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LeetCode-500-键盘行

题目描述：给你一个字符串数组 words ，只返回可以使用在 美式键盘 同一行的字母打印出来的单词。

示例说明请见LeetCode官网。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/probl...
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解法一：遍历字符

首先，如果words为null或者words为空数组，则直接返回空数组，就是没有符合条件的单词。

否则，首先初始化一个Map为characterMap用于保存行数和对应行所有的字符，然后遍历words中的单词word：

  • 首先获取word中的第一个字符为firstCharacter，根据characterMap判断第一个字符firstCharacter在哪一行并记录rowNum；
  • 然后，判断word后面的字符是否都在第rowNum行，如果不是，则跳过处理下一个单词；如果是，则把单词添加到结果集里面。

最后返回结果集中的单词。

import java.util.*;

/**
 * @Author: ck
 * @Date: 2021/10/3 10:47 上午
 */
public class LeetCode_500 {
    private static final Map<Integer, Set<Character>> characterMap = new HashMap<>();

    static {
        characterMap.put(1, new HashSet<>(Arrays.asList(new Character[]{'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'})));
        characterMap.put(2, new HashSet<>(Arrays.asList(new Character[]{'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'})));
        characterMap.put(3, new HashSet<>(Arrays.asList(new Character[]{'z', 'x', 'c', 'v', 'b', 'n', 'm'})));
    }

    public static String[] findWords(String[] words) {
        if (words == null || words.length == 0) {
            return new String[0];
        }
        String[] result = new String[words.length];
        int index = 0, size = 0;
        for (String word : words) {
            if (word == null || word.length() == 0) {
                result[index++] = word;
                size++;
                continue;
            }
            char[] wordArr = word.toCharArray();
            char firstCharacter = wordArr[0];
            int rowNum = 1;
            for (Map.Entry<Integer, Set<Character>> characterEntry : characterMap.entrySet()) {
                if (characterEntry.getValue().contains(Character.toLowerCase(firstCharacter))) {
                    rowNum = characterEntry.getKey();
                    break;
                }
            }
            int i;
            for (i = 1; i < word.length(); i++) {
                if (!characterMap.get(rowNum).contains(Character.toLowerCase(wordArr[i]))) {
                    break;
                }
            }
            if (i == word.length()) {
                result[index++] = word;
                size++;
            }
        }
        return Arrays.copyOf(result, size);
    }

    public static void main(String[] args) {
        String[] words = new String[]{"Hello", "Alaska", "Dad", "Peace"};
        for (String word : findWords(words)) {
            System.out.println(word);
        }
    }
}

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