LeetCode58: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.

很自然想到从字符串尾端开始遍历直到找到第一个为’ ‘的字符即可，但是需要注意到这样的测试用例”aa “，即字符串最右端为空字符，这是需要注意先去除掉右边的空字符再开始计数。

runtime:4ms

class Solution {
public:
    int lengthOfLastWord(string s) {
        int length=s.size();
        int result=0;
        int i=length-1;

        //首先找到右边第一个不为‘ ‘的下标
        while(i>=0&&s[i]==‘ ‘) i--;

        //开始遍历字符串，找到第一个为‘ ‘的即可停止
        for(;i>=0;i--)
        {
            if(s[i]!=‘ ‘)
            {
                result++;
            }
            else
            {
                break;
            }
        }
        return result;
    }
};