LeetCode: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.

解法一：STL

class Solution {
public:
    int lengthOfLastWord(string s) {
        //STL
        auto first=find_if(s.rbegin(),s.rend(),::isalpha);
        auto last=find_if_not(first,s.rend(),::isalpha);
        return distance(first,last);
    }
};

解法二：

class Solution {
public:
    int lengthOfLastWord(string s) {
        //从后往前扫描
        
        int len=s.size();
        //去掉后缀空格
        
        while(s[len-1]==' ')
            len--;
        
        int sum=0;
        
        for(int i=len-1;i>=0;i--)
        {
            if(s[i]!=' ')
                sum++;
            else 
                break;
        }
        return sum;
        
    }
};

解法三：顺序扫描

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