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; }…

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; } };…

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…

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…