LeetCode 58. Length of Last Word
阿新 • • 發佈:2017-12-22
per 這樣的 int last -c efi empty ets !=
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.
Example:
Input: "Hello World" Output: 5
這道題簡直太簡單,面試都考這樣的就好了,需要求最後一個單詞的長度,如果不存在最後返回0。而且s只存在字母和空格,不存在其他的字符,只需要從後向前遍歷數字符就可以了,代碼如下:
1 class Solution { 2 public: 3 int lengthOfLastWord(string s) 4 { 5 int len = 0, tail = s.length() - 1; 6 while (tail >=0 && s[tail] == ‘ ‘) 7 tail--; 8 while(tail >=0 && s[tail] != ‘ ‘) 9 { 10 tail--; 11 len++; 12 } 13 return len; 14 } 15 };
LeetCode 58. Length of Last Word