leetCode 58. Length of Last Word 字符串

标签：字符串

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.

class Solution {
public:
    vector<string> stringSplit(string s, const char * split)
    {
        vector<string> result;
        const int sLen = s.length();
        char *cs = new char[sLen + 1];
        strcpy(cs, s.data());
        char *p;
     
        p = strtok(cs, split);
        while (p)
        {
            printf("%s\n", p);
            string tmp(p);
            result.push_back(tmp);
            p = strtok(NULL, split);
        }
        return result;
    }
    int lengthOfLastWord(string s) {
        if(s.size() == 0)
            return 0;
        vector<string> words = stringSplit(s," ");
        if(words.size() == 0)
            return 0;
        return words[words.size() - 1].size();
    }
};

2016-08-10 18:11:30

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leetCode 58. Length of Last Word 字符串

标签：字符串

原文地址：http://qiaopeng688.blog.51cto.com/3572484/1836629