leetcode500: Keyboard Row

最新推荐文章于 2021-01-24 22:56:51 发布

UESTC_kb216

最新推荐文章于 2021-01-24 22:56:51 发布

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本文介绍了一种算法，用于从给定的字符串数组中筛选出仅使用美国键盘单行字母排列即可输入的单词。通过将键盘上的三行字母分别存储为集合，并检查每个单词的所有字符是否属于同一行来实现。

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Given a List of words, return the words that can be typed using letters of alphabet on only one row's of American keyboard like the image below.

American keyboard

Example 1:

Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]

Note:

You may use one character in the keyboard more than once.
You may assume the input string will only contain letters of alphabet.

注意：这个程序说明如何返回一个String[]

	public String[] findWords(String[] words) {
		String q1 = "QWERTYUIOP";
		String a1 = "ASDFGHJKL";
		String z1 = "ZXCVBNM";
		ArrayList q = new ArrayList();
		ArrayList a = new ArrayList();
		ArrayList z = new ArrayList();
		List<String> re = new LinkedList<>();
		for (int i = 0; i < q1.length(); i++)
			q.add(q1.charAt(i));
		for (int i = 0; i < a1.length(); i++)
			a.add(a1.charAt(i));
		for (int i = 0; i < z1.length(); i++)
			z.add(z1.charAt(i));
		for (int i = 0; i < words.length; i++) {
			char[] candidate = words[i].toUpperCase().toCharArray();
			boolean flag1 = true;
			boolean flag2 = true;
			boolean flag3 = true;
			for (int j = 0; j < candidate.length; j++) {
				if (q.contains(candidate[j]))
					flag1 = false;
				else if (a.contains(candidate[j]))
					flag2 = false;
				else if (z.contains(candidate[j]))
					flag3 = false;
				else
					break;
			}
			if (((flag1 == false) && (flag2 == true) && (flag3 == true))
					|| ((flag1 == true) && (flag2 == false) && (flag3 == true))
					|| ((flag1 == true) && (flag2 == true) && (flag3 == false)))
				re.add(words[i]);
		}
		return re.toArray(new String[0]);
	}