題目鏈接

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.

---

?

借助strlen函數，相當于遍歷兩遍字符串 ? ? ? ? ? ? ? ? ? ? ? 本文地址

1 class Solution { 2 public: 3 int lengthOfLastWord(const char *s) { 4 if(s == NULL)return 0; 5 int len = strlen(s), i, res = 0; 6 for(i = len-1; i >= 0 && s[i] == ' '; i--);//從尾部開始找到第一個非空格字符 7 for(; i >= 0 && s[i] != ' '; i--)res++; 8 return res; 9 } 10 };

遍歷一遍字符串

1 class Solution { 2 public: 3 int lengthOfLastWord(const char *s) { 4 if(s == NULL)return 0; 5 int res = 0, cnt = 0; 6 for(; *s != '\0'; s++) 7 { 8 if(*s == ' ') 9 { 10 if(cnt != 0) 11 res = cnt; 12 cnt = 0; 13 } 14 else cnt++; 15 } 16 return cnt == 0 ? res : cnt; 17 } 18 };

【版權聲明】轉載請注明出處：http://www.cnblogs.com/TenosDoIt/p/3722115.html

?

?

?

轉載于:https://www.cnblogs.com/TenosDoIt/p/3722115.html

總結

以上是生活随笔為你收集整理的LeetCode:Length of Last Word的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。