[LeetCode] Length of Last Word 字符串查找

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.

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String

 

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    这题比较简单，只是可能前面有空格，后面有空格。- -

算法逻辑：

1. 排除最后的空格
2. index 从后往前查第一个空格。
3. 返回长度。

 #include <iostream>
 #include <cstring>
 using namespace std;

 class Solution {
 public:
     int lengthOfLastWord(const char *s) {
         int n = strlen(s);
         if(n <)    return ;
         int idx=n-;
         while(idx>=&&s[idx]==' ')  idx--;
         n = idx+;
         while(idx>=){
             if (s[idx]==' ')  break;
             idx --;
         }
 //        if(idx<0)   return 0;
         return n - idx -;
     }
 };

 int main()
 {
     char s[] = " 12   ";
     Solution sol;
     cout<<sol.lengthOfLastWord(s)<<endl;
     return ;
 }