第十三周OJ-Q58解题方法

问题：

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.

这个要求最后一个单词的长度，那我们可以从后找起，跳过所有的空格，找到这个句子真正的尾巴。当然，这一点是被这个奇葩的测试用例 a（空格）恶心到了才想到的。然后往前扫，直到碰到另一个空格。其间每走一步用个变量自增一下。比较简单，直接放代码：

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.