520. Detect Capital【easy】
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.
解法一：
1 class Solution { 2 public: 3 bool detectCapitalUse(string word) { 4 if (word.length() <= 1) { 5 return true; 6 } 7 8 bool flag = false; 9 for (int i = 1; i < word.length(); ++i) { 10 //以小写字母开头，则后面都必须为小写 11 if (word[0] >= 'a' && word[0] <= 'z') { 12 if (word[i] < 'a' || word[i] > 'z') { 13 return false; 14 } 15 } 16 //以大写字母开头，下面分类讨论 17 else if (word[0] >= 'A' && word[0] <= 'Z') { 18 if (i == 1 && word[i] >= 'a' && word[i] <= 'z') { 19 flag = true; 20 continue; 21 } 22 //说明后面都必须为小写 23 if (flag) { 24 if (word[i] >= 'A' && word[i] <= 'Z') { 25 return false; 26 } 27 } 28 //说明后面都必须为大写 29 else { 30 if (word[i] >= 'a' && word[i] <= 'z') { 31 return false; 32 } 33 } 34 } 35 } 36 37 return true; 38 } 39 };
解法二：
1 class Solution { 2 public: 3 bool detectCapitalUse(string word) { 4 int size=word.size(),count=0; 5 if(size<=1) 6 return true; 7 for (int i = 1; i < size; i++){ 8 if(word[i]>='a'&&word[i]<='z') 9 count+=1; 10 else 11 count+=2; 12 } 13 if(count==size-1) 14 return true; 15 else if(count==2*(size-1)) 16 return word[0]>='A'&&word[0]<='Z'; 17 else 18 return false; 19 } 20 };
参考@zhengchaojie 的代码。
From 1~size-1,if we meet with a-z,we add 1,else we add 2.Then we can get the result that if the second to last letter is all lowercase or all upcase.
解法三：
1 bool detectCapitalUse(string word) { 2 const char *c = word.c_str(); 3 if (word.size() <= 1) return true; 4 if (*c <= 'z' && *c >= 'a') { 5 c = c + 1; 6 while (*c) { 7 if (*c <= 'Z' && *c >= 'A') return false; 8 c = c + 1; 9 } 10 } else { 11 c = c + 1; 12 if (*c <= 'Z' && *c >= 'A') { 13 c = c + 1; 14 while (*c) { 15 if (*c <= 'z' && *c >= 'a') return false; 16 c = c + 1; 17 } 18 } else { 19 c = c + 1; 20 while (*c) { 21 if (*c <= 'Z' && *c >= 'A') return false; 22 c = c + 1; 23 } 24 } 25 } 26 27 return true; 28 }
参考@ai977313677 的代码。
解法四：
1 class Solution { 2 public: 3 bool detectCapitalUse(string word) { 4 int cnt = 0; 5 for (char c : word) if (isupper(c)) ++cnt; 6 return !cnt || cnt == word.size() || cnt == 1 && isupper(word[0]); 7 } 8 };
参考@lzl124631x 的代码。