LeetCode: 520. Detect Capital

最新推荐文章于 2021-01-27 20:38:49 发布

原创最新推荐文章于 2021-01-27 20:38:49 发布 · 246 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

LeetCode 专栏收录该内容

35 篇文章

订阅专栏

本文介绍了一个LeetCode上的编程题目——检测字符串中大写字母的使用是否符合规范。提供了三种解决方案，包括一个简洁的正则表达式匹配方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

LeetCode: 520. Detect Capital

Given a word, you need to judge whether the usage of capitals in it is
right or not.

We define the usage of capitals in a word to be right when one of the
following cases holds:

All letters in this word are capitals, like “USA”. All letters in this
word are not capitals, like “leetcode”. Only the first letter in this
word is capital if it has more than one letter, like “Google”.
Otherwise, we define that this word doesn’t use capitals in a right
way. Example 1: Input: “USA” Output: True Example 2: Input: “FlaG”
Output: False Note: The input will be a non-empty word consisting of
uppercase and lowercase latin letters.

自己的答案：

public class Solution {
    public boolean detectCapitalUse(String word) {
        if (word == null) {
            return false;
        }
        int length = word.length();
        if (length == 1) {
            return true;
        }
        String first = word.substring(0, 1);
        String temp = word.substring(1);
        if (temp.toLowerCase().equals(temp)) {
            return true;
        }
        if (first.toUpperCase().equals(first)) {
            if (temp.toUpperCase().equals(temp)) {
                return true;
            }
        }
        return false;
    }
}

最快的答案：

public class Solution {
    public boolean detectCapitalUse(String word) {

        boolean flag = true;

        if (((int)word.charAt(0) >64)&&((int)word.charAt(0) <=90)){
            if (word.length()>=2){
                if (((int)word.charAt(1) >64)&&((int)word.charAt(1) <=90))
                    for (int i =2; i < word.length(); i++){
                        if (!(((int)word.charAt(i) >=65)&&((int)word.charAt(i) <=90)))
                            flag = false;

                    }


                else{
                    for (int i =1; i < word.length(); i++){
                            if ((((int)word.charAt(i) >=65)&&((int)word.charAt(i) <=90)))
                                flag = false;

                        }
                }
            }
        }
        else{
            for (int x =0; x < word.length(); x++){
                    if ((((int)word.charAt(x) >64)&&((int)word.charAt(x) <=90)))
                        flag = false;
                }
        }

        if (flag)
            return true;
        else 
            return false;

    }
}

最简洁的答案：

public boolean detectCapitalUse(String word) {
    return word.matches("[A-Z]*|[A-Z]?[a-z]*");
}

Simple Java Solution O(n) time O(1) space