<LeetCode OJ> 58. Length of Last Word

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

.

分析：

简单模拟思想。去掉尾部的空格并统计是字符的个数直到再次遇到空格就停止统计

class Solution {
public:
int lengthOfLastWord(string s) {
int strLen=s.size();
int len=0;
int i=0;
while(s[strLen-i-1]==' ')
i++;
for(;i<strLen;i++)
{
if(s[strLen-i-1]>='A'&&s[strLen-i-1]<='Z' || s[strLen-i-1]>='a'&&s[strLen-i-1]<='z' )
len++;
else
break;
}
return len;
}
};

或者用STL来写：

class Solution {
public:
int lengthOfLastWord(string s) {
int i = s.find_last_not_of(' ');
return (i == string::npos) ? 0 : (i - s.find_last_of(' ', i));
}
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50498956

原作者博客：http://blog.csdn.net/ebowtang