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 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…

原题地址:https://oj.leetcode.com/problems/length-of-last-word/ 题意: 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 wo…

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…

Length of Last Word Total Accepted: 47690 Total Submissions: 168587     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. N…

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; } };…

class Solution { public: int lengthOfLastWord(const char *s) { ; string snew=s; ,len=strlen(s); ]; ){ *p--; len--; } ){ n++; p--; len--; } return n; } }; 题意:给一个数组,以空格来判断是否字的结束与开始.一个词的前后都是空格,中间无空格.返回最后一个字的长度. 思路:给的是一个不可修改的字符串,为了方便,创建另外的指针.若最后是空格,先将后面的…