LeetCode 58. Length of Last Word

58. Length of Last Word

一、问题描述

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-space characters only.

二、输入输出

For example,
Given s = "Hello World",
return 5.

三、解题思路

• 有种情况不知道你有没有想到，就是前面是单词，末尾是空格。反正我刚开始是漏掉了。。
• 很简单，string.find_last_not_of找到最后一个不是空格的位置，如果没有找到，说明全是空格，或者是空串，返回0；否则就找到最后一个单词的末尾位置。然后从这个位置开始向前遍历，直到遇到空格或者到大字符串开头停止，即可计算出最后一个单词的长度。

class Solution {public:    int lengthOfLastWord(string s) {        size_t found = s.find_last_not_of(' ');        if(found == string::npos)return 0;        int i = found;        while(i >= 0 && s[i] != ' ') i--;        if(i < 0)return found+1;        else return (found - i);    }};