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 cha…

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 cha…

［抄题］: [暴力解法]: 时间分析: 空间分析: [优化后]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: "b a " 最后一位是空格,可能误判lastindexof().所以必须用.trim() ［思维问题］: ［一句话思路］: 用函数 再次强调是最后一位的索引是length() - 1 ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: ［二刷］: ［三刷］: ［四刷］: ［…

找出最后一个词的长度 class Solution { public: int lengthOfLastWord(string s) { , l = , b = ; while((b = s.find(" ", a)) != string::npos){ ) l = b - a; a = b + ; } if(a == s.size()) return l; else return s.size() - a; } };…

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 char…

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 cha…

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. Note: A word is defined as a character sequence consists of non-spa…

Length of Last Word  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 cons…

题目: 最后一个单词的长度 给定一个字符串, 包含大小写字母.空格' ',请返回其最后一个单词的长度. 如果不存在最后一个单词,请返回 0 . 样例 给定 s = "Hello World",返回 5. 注意 一个单词的界定是,由字母组成,但不包含任何的空格. 解题: 利用正则确认是字母,向前走,当遇到不是字母的时候跳出程序,为了合适的跳出定义一个布尔值,是字母的时候是true,当不是字母同时布尔值是true时候跳出 Java程序: public class Solution { /*…

给定一个字符串, 包含大小写字母.空格 ' ',请返回其最后一个单词的长度.如果不存在最后一个单词,请返回 0 .注意事项:一个单词的界定是,由字母组成,但不包含任何的空格.案例:输入: "Hello World"输出: 5详见:https://leetcode.com/problems/length-of-last-word/description/ Java实现: class Solution { public int lengthOfLastWord(String s) { in…

1. 原题链接 https://leetcode.com/problems/length-of-last-word/description/ 2. 题目要求 给定一个String类型的字符串,字符串中可能含有空格‘  ’.从字符串的末尾开始查找,找到第一个单词,该单词中间不能有空格,并返回其长度. 3. 解题思路 首先判断该字符串是否为空,为空直接返回 0: 否则,从尾部向前查找,使用两个while循环,第一个while循环用于过滤空格,当该位置不为空格时,跳出while循环: 然后进入第二个w…

https://leetcode.com/problems/length-of-last-word/ int lengthOfLastWord(char* s) { int ans = 0; int fans = 0; for(int i = 0;s[i];i++){ if (s[i] ==' '){fans = ans;ans = 0;while(s[i + 1] == ' '){i++;}} else ans++; } return ans?ans:fans; }…

1.题目 58. Length of Last Word——Easy 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 la…

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 cha…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 库函数 双指针 单指针 日期 题目地址:https://leetcode.com/problems/length-of-last-word/description/ 题目描述 Given a string s consists of upper/lower-case alphabets and empty space characters ' ',…

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 cha…

一天一道LeetCode系列 (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 cons…

[ 问题: ] 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. 给你一个字符串,设法获取它最后一个单词的长度.假设这个单词不存在,则返回0. [ 分析 : ] A word is defined…

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 cha…

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 cha…

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 cha…

leetcode: 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…

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…

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 cha…

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 cha…

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…

这个题目很简单,给一个字符串,然后返回最后一个单词的长度就行.题目如下: 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 charact…

题目简述 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-spac…

题目: 输入字符串 s,返回其最后一个单词的长度 如 s="Hello World"   返回5 s="Hello World    "   返回5 s="  "     返回0 开始从前向后判断,超时了.改成从后向前判断,通过了. class Solution { public: int lengthOfLastWord(const char *s) { ; int slen = strlen(s); ; i >= ; i--) { if…

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 cha…