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LeetCode:Length of Last Word

时间:2014-06-26 13:55:01      阅读:179      评论:0      收藏:0      [点我收藏+]

标签:leetcode

     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.


解题报告:

   

    用个变量last保存遍历到当前位置的最后一个空格位置,只要当前字符不是空格,就更新


res,这样下来,最终res里面存放的就是所要求的结果.需要注意的是,last变量的初值.


class Solution {
public:
    int lengthOfLastWord(const char *s) 
    {
        if (s == NULL)
            return 0;
        int res = 0,last = -1;
        for (int i = 0; s[i]; ++i)
        {
            if( s[i] == ' ')
            {
                last = i;
                continue;
            }
            res = i - last;
        }
        return res;
    }
};


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LeetCode:Length of Last Word

标签:leetcode

原文地址:http://blog.csdn.net/dream_you_to_life/article/details/34471409

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