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 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…

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…

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…

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…

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…

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…

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…

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…

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…

[ 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…

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…

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…

［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］: ［…

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. N…

#-*- coding: UTF-8 -*-#利用strip函数去掉字符串去除空格(其实是去除两边[左边和右边]空格)#利用split分离字符串成列表class Solution(object):    def lengthOfLastWord(self, s):        """        :type s: str        :rtype: int        """        if s==None:return 0     …

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…