}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 OJ:Length of Last Word(最后一个词的长度) Given a stringsconsists 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 ...
函数的接口决定了字符串的长度是未知的,要自己循环找一下,然后从尾向头找不等于空格的字符,找到了就找到了最后一个单词,然后累计直到空格结束。 class Solution { public: int lengthOfLastWord(const char *s) { if(s == NULL) return 0; int len = 0, res = 0; while(s[len] != '\0') len++...
LeetCode Length of Last Word https://leetcode.com/problems/length-of-last-word/ 题目: Given a stringsconsists 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...
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:
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
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. ...
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" ...
public int lengthOfLastWord(String s) { int idx = s.length() - 1; // 跳过末尾的空格 while(idx >= 0){ if(s.charAt(idx) != ' ') break; idx--; } // 记录结束位置 int end = idx; // 如果已经超界返回0 if(idx < 0) return 0; ...
Introduce 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...LeetCode 58:Length of Last Word Given a string s consists of upper/lower-case alpha...