LeetCode-58. Length of Last Word

最新推荐文章于 2024-09-20 00:30:00 发布

零件儿

最新推荐文章于 2024-09-20 00:30:00 发布

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本文链接：https://blog.youkuaiyun.com/weixin_40744915/article/details/79699961

本文提供了一种解决LeetCode上“最后一个单词的长度”问题的方法，通过Python实现，展示了如何有效地找到并计算字符串中最后一个单词的长度，包括去除尾部空白字符及识别单词边界的技术。

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评论区最多的就是it's no fun at all!哈哈哈

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

solution：

class Solution:
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
if s is None:
return 0
j = s.strip().split(' ')

return len(j[-1])

当然也有那么一些大佬愿意不用函数：

用==' '或者.isspace()来判断最后一个空格后的那个单词的长度：

def lengthOfLastWord(self, s):
    ls = len(s)
    # slow and fast pointers
    slow = -1
    # iterate over trailing spaces
    while slow >= -ls and s[slow] == ' ':
        slow-=1
    fast = slow
    # iterate over last word
    while fast >= -ls and s[fast] != ' ':
        fast-=1
    return slow - fast

def lengthOfLastWord(self, s):
    # (going backwards) find first non-space char
    for i in range(len(s) - 1, -2, -1):
        if i == -1 or s[i] != " ":
            break
    
    # keep going until a space or end of string
    for j in range(i, -2, -1):
        if j == -1 or s[j] == " ":
            return i - j