leetcode-58-Length-of-Last-Word

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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.

分析

求最后一个单词的字母数。如果不用 C++ 的 STL 的话，还是得将 string 转换为 c_str，也就是 char *，所以，试试纯粹的C？

解决方案1(C)

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int lengthOfLastWord(char* s) {
    if(!s) {
        return 0;
    }

    int result = 0;
    int slen = strlen(s);
    while(s[slen-1] == ' ') {
        slen--;
    }
    for(int i = slen-1; i>=0; i--) {
        if(s[i] == ' ') {
            break;
        }
        result++;
    }
    return result;
}

解决方案1(C++)

如果不用STL，可以将string 类型转换为C的c_str

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class Solution {
public:
    int lengthOfLastWord(string s) {
        return lengthOfLastWord(s.c_str());
    }
private:
    int lengthOfLastWord(const char* s) {
        if(!s) {
            return 0;
        }

        int result = 0;
        int slen = strlen(s);
        while(s[slen-1] == ' ') {
            slen--;
        }
        for(int i = slen-1; i>=0; i--) {
            if(s[i] == ' ') {
                break;
            }
            result++;
        }
        return result;
    }
};

解决方案3(Python)

最简洁的还是Python

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class Solution(object):
    def lengthOfLastWord(self, s):
        """
        :type s: str
        :rtype: int
        """
        s = s.strip()
        word_list = s.split(" ")
        try:
            last_word = word_list[-1]
            return len(last_word)
        except IndexError:
            return 0

题目来源