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C++
topic: 58. 最后一个单词的长度 - 力扣(LeetCode)
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Inspect the topic at the begining.
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This topic seems very easy. And today I really want to talks something about Being fucking Professional. I I've been watching the TV series 李狗嗨 recently. Fighting for your client's interests all over, that's all we lawyers can do.
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And flash back to the topic.
Let's travel from the end. if s[i] = ' ', then end loop.
class Solution {
public:
int lengthOfLastWord(string s) {
int n = s.size(); // 获取字符串长度
int count = 0; // 用于记录最后一个单词的长度
// 从字符串末尾开始遍历
for (int i = n - 1; i >= 0; i--)
{
// 如果当前字符不是空格,开始计数
if (s[i] != ' ')
{
count++;
// 如果遇到空格或者到达字符串开头,跳出循环
if (i == 0 || s[i - 1] == ' ')
{
break;
}
}
}
return count;
}
};