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.
这道题给了我们一个单词，让我们检测大写格式是否正确，规定了三种正确方式，要么都是大写或小写，要么首字母大写，其他情况都不正确。那么我们要做的就是统计出单词中所有大写字母的个数cnt，再来判断是否属于这三种情况，如果cnt为0，说明都是小写，正确；如果cnt和单词长度相等，说明都是大写，正确；如果cnt为1，且首字母为大写，正确，其他情况均返回false，参见代码如下：
解法一：
class Solution {
public:
bool detectCapitalUse(string word) {
int cnt = , n = word.size();
for (int i = ; i < n; ++i) {
if (word[i] <= 'Z') ++cnt;
}
return cnt == || cnt == n || (cnt == && word[] <= 'Z');
}
};
下面这种方法利用了STL的内置方法count_if，根据条件来计数，这样使得code非常简洁，两行就搞定了，丧心病狂啊～
解法二：
class Solution {
public:
bool detectCapitalUse(string word) {
int cnt = count_if(word.begin(), word.end(), [](char c){return c <= 'Z';});
return cnt == || cnt == word.size() || (cnt == && word[] <= 'Z');
}
};
参考资料：
https://discuss.leetcode.com/topic/79912/3-lines
https://discuss.leetcode.com/topic/79930/java-1-liner
https://discuss.leetcode.com/topic/80314/6ms-2-lines-c-solution/2
https://discuss.leetcode.com/topic/79911/simple-java-solution-o-n-time-o-1-space