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题目描述：(链接)
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.
解题思路：
简单题，注意去掉字符串尾部的空格！！
1 class Solution { 2 public: 3 int lengthOfLastWord(string s) { 4 int len = s.size(); 5 int j = len - 1; 6 while (s[j] == ‘ ‘) { --j; } 7 8 int index = 0; 9 for (int k = 0; k < j; ++k) { 10 if (s[k] == ‘ ‘) { 11 index = k + 1; 12 } 13 } 14 15 return j - index + 1; 16 } 17 };
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原文地址：http://www.cnblogs.com/skycore/p/4936308.html
