1. 问题描述

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.

Tags: String

2. 解题思路

• 字符串操作

3. 代码

class Solution {
public:
    int lengthOfLastWord(string s) {
        string::size_type pos = s.find_last_not_of(‘ ‘);
        if (pos != s.npos)
        {
            string sTemp = s.substr(0, pos+1);
            pos = sTemp.rfind(‘ ‘);
            if (pos != sTemp.npos)
            {
                return sTemp.length() - pos - 1;
            }
            else
            {
                return sTemp.length();
            }
        }
        return 0;
    }
};

4. 反思