(C++練習) 58. Length of Last Word
阿新 • • 發佈:2019-02-23
NPU upper pty style The world per 字串 lower
題目 :
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
大意 :
找出最後一個字串的長度, 利用space characters 來判斷
解法 : 直接從最後面開始計數字串的長度, 遇到space即回傳長度
1 class Solution { 2 public: 3 int lengthOfLastWord(string s) { 4 5 int len = strlen(s.c_str()) - 1; 6 7 int ans = 0; 8 for (int i = len; i >= 0; i--){ 9 if (s[i] != ‘‘) ans++; 10 else if (ans) return ans; 11 } 12 13 return ans; 14 } 15 };
(C++練習) 58. Length of Last Word