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.

这道题难度不是很大。先对输入字符串做预处理，去掉开头和结尾的空格，然后用一个计数器来累计非空格的字符串的长度，遇到空格则将计数器清零。代码如下：

class Solution {public:    int lengthOfLastWord(const char *s) {        int len = strlen(s);        int left = 0;        int right = len - 1;        int count = 0;        while (s[left] == ‘ ‘) ++left;        while (s[right] == ‘ ‘) --right;        for (int i = left; i <= right; ++i) {            if (s[i] == ‘ ‘) count = 0;            else ++count;        }        return count;    }};

Length of Last Word 求末尾单词的长度