LeetCode 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.

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解法一：参考别人的解法,太过于巧妙，非常棒的解法

 int lengthOfLastWord( char* s) {        int len = 0;        while (*s) {            if (*s++ != ' ')                ++len;            else if (*s && *s != ' ')                len = 0;        }        return len;    }

解法二：自己的解法，比较容易理解

 int lengthOfLastWord(string s) {        int len=s.size();        for (int i=0;i<len;) {            int count=0;            while (isalpha(s[i])&&i<len) {                count++;                i++;            }             while (isspace(s[i])&&i<len)                i++;            if (i==len)                return count;        }        return 0;    }