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.

题很简单，直接上代码吧！

代码如下：

public class Solution{    public int lengthOfLastWord(String s)    {        if(s==null||s.length()<=0)            return 0;        s=s.trim();        if(s.length()<=0)            return 0;        int res=(s.length()-1) - (s.lastIndexOf(" ")+1) +1 ;        return res;    }}

下面是C++做法

代码如下：

#include <iostream>#include <vector>#include <string>using namespace std;class Solution {public:    int lengthOfLastWord(string s)     {        int i = 0;        for (i = s.length() - 1; i >= 0; i--)        {            if (s[i] != ' ')                break;        }        int count = 0;        for (; i >= 0; i--)        {            if(s[i]==' ')                break;            else                count++;        }        return count;    }};