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[LeetCode] Length of Last Word 字符串查找

2024-08-06 00:41:41   216人阅读

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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[LeetCode] Length of Last Word 字符串查找


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