题目源自于leetcode。简单题。
题目:Given a string s consists
of upper/lower-case alphabets and empty space characters ' ',
return the length of last word in the string.
If the last word does not exist, return 0.Note: A word is defined as a character sequence consists of non-space characters only.For example, Given s = "Hello
World",return 5.
思路:从右向左。先排除字符串末尾的空格。再找到单词的长度。
代码:
class Solution {
public:
int lengthOfLastWord(const char *s) {
int n = strlen(s);
int len = 0;
n--;
while(n >= 0 && s[n] == ' ')
n--;
while(n >= 0 && s[n] != ' ')
{
n--;
len++;
}
return len;
}
};

本文详细介绍了如何通过C++代码解决LeetCode中的简单问题,即计算给定字符串中最后一个单词的长度。代码从右向左遍历字符串,首先排除末尾的空格,然后计算单词的长度。
490

被折叠的 条评论
为什么被折叠?



