Question :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 non-space characters only. For example, Given...
Given a string s consisting of words and spaces, returnthe length of thelastword in the string. 给出一个字符串s,由若干单词组成,单词前后用一些空格字符隔开。返回字符串中最后一个单词的长度。 Awordis a maximal substring consisting of non-space characters only. 单词是指仅由字母组成、不包含任何空...
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 non-space characters only. Example: Input: "Hello Worl...
Length of Last Word 最后个字符长度 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: For example, Given s = ...
#58.Length of Last Word先尽量正确理解题目: 给定一个字符串,其中包括了大小写敏感的字母组和空字符' ',返回最后一个单词的长度(从左往右顺序的最有一个单词) 注意:何为一个单词Word,即不包含空字符的最大子字符串。 Examples: Input: "Hello World", Output: 5. 因为最后一个单词World长度为5. ...
Leetcode 之Length of Last Word(37),扫描每个WORD的长度并记录即可。intlengthOfLast(constchar*s){//扫描统计每个word的长度intlen=0;while(*s){if(*s++!='')//注意不管是否满足s都要++len++;elseif(*s&&*s
简单题,其实题目假设了不会出现数字字符等,不然这样做是过不了的。还需要判断是否这个word全为65 <= ord() <= 122 代码语言:javascript 复制 classSolution(object):deflengthOfLastWord(self,s):""":type s:str:rtype:int"""returnlen(s.strip().split(" ")[-1]) ...
There will be at least one word in s. 思路: 这道题要求我们求字符串最后一个单词的长度,很简单倒序遍历,遇到空格跳过,直到遇到第一个字母,开始计数,再遇到空格就退出即可,代码如下: class Solution{public:intlengthOfLastWord(string s){if(s.empty())return0;if(s.size()==1)return1;intcoun...
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. Example: Input: "Hello World" Output: 5 Note: A word is defined as a character sequence consists of non-space...
int lengthOfLastWord(const char *s) { int result = 0; bool flag = false; while(*s != '\0') { if(*s == ' ') flag = true; else if(flag == true) { flag = false; result = 1; } else Example 7Source File: length-of-last-word.cc From epicode with MIT License 5 votes...