LeetCode Easy: 58. Length of Last Word

一、题目

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.

Example:

Input: "Hello World"
Output: 5

输出给定字符串的最后一个单词的长度

二、解题思路

我首先想到的就是从后往前遍历给定字符串，直到遇到空格位置，计数即可，AC之后，查看了网上的做法，比我简单多了，利用的是python中的切片功能split（）方法。

三、代码

#coding:utf-8
def lengthOfLastWord1(s):
    """
    :type s: str
    :rtype: int
    """
    if s == " ":
        return 0
    if len(s) == 1:
        return 1
    j = len(s)
    while s[j-1] != " ":
        j-=1
        # if j == 0:
        #     return
    count = len(s) - j
    print(count)
    return count

def lengthOfLastWord2(s):
    tmp = s.split()
    if not len(tmp):
        return 0
    else:
        print(len(tmp[-1]))
        return len(tmp[-1])

if __name__ == '__main__':
    s = "IndexError: string index out of range"
    lengthOfLastWord2(s)
    # print(s.split('i'))
    # print(s.split('i',1))