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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.
思路：从后向前遍历字符串，先过滤掉空格，直到遇到空格停止。
1 class Solution 2 { 3 public: 4 int lengthOfLastWord(string s) 5 {//返回s中最后一个单词的长度 6 if(s.size() == 0) 7 return 0; 8 9 int end = s.size()-1; 10 while(end >=0 && s[end] == ‘ ‘) 11 end--;//过滤掉后面的空格符 12 int i=end; 13 while(i>=0) 14 {//从后向前遍历 15 if(s[i] == ‘ ‘) 16 break; 17 i--; 18 } 19 20 return end-i; 21 } 22 }; 23 int main() 24 { 25 Solution sol; 26 string test[] = {"Hello world","Hello world ",""," "}; 27 28 for(auto const &i : test) 29 cout << sol.lengthOfLastWord(i)<<endl; 30 return 0; 31 }
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原文地址：http://www.cnblogs.com/dreamrun/p/4546732.html
