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:

  1. All letters in this word are capitals, like "USA".
  2. All letters in this word are not capitals, like "leetcode".
  3. 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.

很简单的一道题,渣渣如我只能暴力求解:

1.brutal force/命令式编程

 public class Solution {
public boolean detectCapitalUse(String word) {
char[] value = word.toCharArray();
if(value.length == 1 || word == null) {
return true;
}
boolean secIsUC = false;
if(Character.isLowerCase(value[0])) {
for(int i=1; i<value.length; i++) {
if(Character.isUpperCase(value[i])) {
return false;
}
}
} else {
secIsUC = Character.isUpperCase(value[1]) ? true : false;
if(secIsUC) {
for(int i=2; i<value.length; i++) {
if(Character.isLowerCase(value[i])) {
return false;
}
}
} else {
for(int i=2; i<value.length; i++) {
if(Character.isUpperCase(value[i])) {
return false;
}
}
}
}
return true;
}
}

2.暴力求解升级版

1 public class Solution {
2 public boolean detectCapitalUse(String word) {
3 int cnt = 0;
4 for(char c: word.toCharArray()) if('Z' - c >= 0) cnt++;
5 return ((cnt==0 || cnt==word.length()) || (cnt==1 && 'Z' - word.charAt(0)>=0));
6 }
7 }

3.声明式编程

 public boolean detectCapitalUse(String word) {
if (word.length() < 2) return true;
if (word.toUpperCase().equals(word)) return true;
if (word.substring(1).toLowerCase().equals(word.substring(1))) return true;
return false;
}

4.RegEx

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

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