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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 consists of non-space characters only.

For example,

Given s = “Hello World”,

return 5.

（二）解题

题目比较简单，从尾向头找，第一个不为空格的字母，然后再继续找遇到空格为止。纪录单词长度。

class Solution {public:    int lengthOfLastWord(string s) {        if(s.length() == 0) return 0;        int count = 0;        int i = s.length()-1;        while(s[i]==' ') i--;//忽略前面的空格        while(i>=0&&s[i]!=' ') //当为字母的时候计算长度，为空格就退出循环        {            count++;            i--;        }        return count;    }};

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