}String[] arr = s.split(" ");returnarr[arr.length-1].length(); } 04 第三种解法 将原字符串首尾的空格去掉,然后找到最后一次出现空格的位置,两者相减再减1即为最后单词的长度。 public intlengthOfLastWord3(String s) { return s.trim().length()-s.trim().lastIndexOf(" ")-1; } 05 小结...
leetcode 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 (last word means the last appearing word if we loop from left to right) in the string. If the last word does not exist, return 0...
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...
1.1 英文题目 Given a string s consists of some words separated by spaces, return the length of the last word in the string. If the last word does not exist, return 0. A word is a maximal substring consisting of non-space characters only. 1.2 中文题目 给你一个字符串 s,由若干单词组成,...
11最后一个单词的长度 Length of Last Word【Python刷LeetCode 力扣】, 视频播放量 55、弹幕量 0、点赞数 1、投硬币枚数 0、收藏人数 0、转发人数 1, 视频作者 一起学AI丶, 作者简介 ,相关视频:16只出现一次的数字Single Number【Python刷LeetCode 力扣】,08移除元素Remo
Can you solve this real interview question? Length of Last Word - Given a string s consisting of words and spaces, return the length of the last word in the string. A word is a maximal substring consisting of non-space characters only. Example 1:
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. My Solution public class Solution { public static int lengthOfLastWord(String s) { ...
LeetCode 58: Length of Last Word 題目描述 題目說明:給一個字串,裡面可能包含英文字母(alphabets) 與空格(space)。回傳最後一個單字(word)的長度。 例如: "Hello World ", 應回傳 5, 因為 World 的長度為 5。 TDD 開始 Step 1, 新增測試用例。長度為2, 不含空格的 word。 s 為"bc" ...
leetcode 58. Length of Last Word 题目描述: 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. No......
public int lengthOfLastWord(String s) { return s.trim().split(" +")[s.trim().split(" +").length - 1].length(); } } 双指针法 复杂度 时间O(N) 空间 O(1) 思路 从后往前看字符串,跳过所有空格后,记下该结束位置,再到下一个空格,再记录一个开始位置,则长度就是结束位置减去开始位置。