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.
解题报告:
用个变量last保存遍历到当前位置的最后一个空格位置,只要当前字符不是空格,就更新
res,这样下来,最终res里面存放的就是所要求的结果.需要注意的是,last变量的初值.
class Solution {
public:
int lengthOfLastWord(const char *s)
{
if (s == NULL)
return 0;
int res = 0,last = -1;
for (int i = 0; s[i]; ++i)
{
if( s[i] == ' ')
{
last = i;
continue;
}
res = i - last;
}
return res;
}
};LeetCode:Length of Last Word,布布扣,bubuko.com
原文:http://blog.csdn.net/dream_you_to_life/article/details/34471409