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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.
这道题比较简单，注意考虑字符串最后的空格以及全为空格的字符串的这几种特殊情况。
代码如下：
1 class Solution { 2 public: 3 int lengthOfLastWord(string s) { 4 if(s.length() == 0) 5 { 6 return 0; 7 } 8 int n = 0; 9 int l = s.length(); 10 for(int i = l-1; i >= 0; i--) 11 { 12 if(n == 0 && s[i] == ‘ ‘) 13 { 14 continue; 15 } 16 if(s[i] != ‘ ‘) 17 { 18 n ++; 19 } 20 else 21 { 22 break; 23 } 24 } 25 return n; 26 } 27 };
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原文地址：http://www.cnblogs.com/shellfishsplace/p/5855874.html
