题目

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.

分析

无

代码

    /**------------------------------------
    *   日期：2015-02-06
    *   作者：SJF0115
    *   题目: 58.Length of Last Word
    *   网址：https://oj.leetcode.com/problems/length-of-last-word/
    *   结果：AC
    *   来源：LeetCode
    *   博客：
    ---------------------------------------**/
    #include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; class Solution { public: int lengthOfLastWord(const char *s) { int len = strlen(s); int i = len - 1; int lastLen = 0; // 去掉空格 while(s[i] == ' '){ --i; }//while while(i >= 0 && s[i] != ' '){ ++lastLen; --i; }//while return lastLen; } }; int main(){ Solution s; char *str = " q f ";//"hello world"; int result = s.lengthOfLastWord(str); // 输出 cout<