[leedcode 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.

哈哈哈。。。坑死。。。

第一遍。。。没考虑空串。。。

第二遍。。。没考虑“a ”。。空格结尾。。。

好了。。。

class Solution {public:    int lengthOfLastWord(const char *s) {        int num = 0;        int len = strlen(s);        while(s[len-1]==' ')            len--;        for(int i=len;i>=0;i--)        {            if(i == 0)                break;            if(s[i-1] != ' ')                num++;            else                break;        }        return num;    }};