[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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    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。

 1 #include <iostream>

 2 #include <cstring>

 3 using namespace std;

 4 

 5 

 6 class Solution {

 7 public:

 8     int lengthOfLastWord(const char *s) {

 9         int n = strlen(s);

10         if(n <1)    return 0;

11         int idx=n-1;

12         while(idx>=0&&s[idx]==' ')  idx--;

13         n = idx+1;

14         while(idx>=0){

15             if (s[idx]==' ')  break;

16             idx --;

17         }

18 //        if(idx<0)   return 0;

19         return n - idx -1;

20     }

21 };

22 

23 int main()

24 {

25     char s[] = " 12   ";

26     Solution sol;

27     cout<<sol.lengthOfLastWord(s)<<endl;

28     return 0;

29 }

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