0058._length_of_last_word.md

58. Length of Last Word

难度: Easy

刷题内容

原题连接

• https://leetcode.com/problems/length-of-last-word/

内容描述

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.

Example:

Input: "Hello World"
Output: 5

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

class Solution(object):
    def lengthOfLastWord(self, s):
        """
        :type s: str
        :rtype: int
        """
        s = s[::-1].strip()
        return s.find(' ') if s.find(' ') != -1 else len(s)

作弊式做法

class Solution(object):
    def lengthOfLastWord(self, s):
        """
        :type s: str
        :rtype: int
        """
        lst = s.split()
        if len(lst) >= 1:
        	return len(lst[-1])
        return 0

split()方法最低可以分0组，split(' ')最低可以分1组

一行解法：
class Solution(object):
    def lengthOfLastWord(self, s):
        """
        :type s: str
        :rtype: int
        """
        return len(s.strip().split(" ")[-1])

简单优化

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        return len(s.split()[-1])

思路 2

去除末尾空格，然后反转字符串，直接统计前非空字符即可

class Solution:
    def lengthOfLastWord(self, s: str) -> int:        
        l = 0
        for v in s.rstrip()[::-1]:
            if v != ' ':
                l += 1
            else:
                break
        return l