[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

这题比较简单，只是可能前面有空格，后面有空格。- -

算法逻辑：

排除最后的空格
index 从后往前查第一个空格。
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 ;

 }

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[LeetCode] Length of Last Word 字符串查找

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