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.
For example,
Given s = “Hello World”,
return 5.
分析
無
代碼
/**------------------------------------
* 日期:2015-02-06
* 作者:SJF0115
* 題目: 58.Length of Last Word
* 網址:https://oj.leetcode.com/problems/length-of-last-word/
* 結果:AC
* 來源:LeetCode
* 博客:
---------------------------------------**/
#include
#include
#include
#include
using namespace std;
class Solution {
public:
int lengthOfLastWord(const char *s) {
int len = strlen(s);
int i = len - 1;
int lastLen = 0;
// 去掉空格
while(s[i] == ' '){
--i;
}//while
while(i >= 0 && s[i] != ' '){
++lastLen;
--i;
}//while
return lastLen;
}
};
int main(){
Solution s;
char *str = " q f ";//"hello world";
int result = s.lengthOfLastWord(str);
// 輸出
cout<
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