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.
代码：
class Solution {
public:
int lengthOfLastWord(string s) {
for ( int i = s.length()-; i >=; --i )
{
if (s[i]==' ')
{
s.erase(s.end()-);
}
else
{
break;
}
}
const size_t len = s.length();
int ret = ;
for ( size_t i = ; i < len; ++i )
{
if (s[i]!=' ')
{
++ret;
continue;
}
if (s[i]==' ')
{
ret = ;
continue;
}
}
return ret;
}
};
tips：
先把后面的空格都去掉。然后从头遍历，最后留下的ret就是最后一个单词的长度。
还有STL的一个做法，明天再看。
==========================================
第二次过这道题，感觉不知道第一次为啥还用erease这种了，直接一个指针从后往前遍历就AC了。
class Solution {
public:
int lengthOfLastWord(string s) {
int i = s.size()-;
while ( i>= ){
if ( s[i]==' ' )
{
--i;
}
else
{
break;
}
}
int ret = ;
while ( i>= )
{
if (s[i]!=' ')
{
++ret;
--i;
}
else
{
break;
}
}
return ret;
}
};