字符串(1)——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: TrueExample 2:
Input: "FlaG"
Output: FalseNote: 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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