作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
目录
- 日期
题目地址:https://leetcode.com/problems/length-of-last-word/description/
题目描述
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.
Examle:
input:
"Hello World"
"Hello World "
"Hello W orld"
"Hello Wo rld"
output:
5
5
4
3
题目大意
计算一个字符串中,最后一个不为空的单词的长度。
解题方法
库函数
使用库函数,方法比较简单,一行代码。
class Solution(object):
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
return len(s.strip().split(' ')[-1])
双指针
使用两个指针,一个指向最后一个字符串的结尾,一个指向最后一个字符串的开头。
Python代码如下:
class Solution(object):
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
N = len(s)
left, right = 0, N - 1
while right >= 0 and s[right] == " ":
right -= 1
left = right
while left >= 0 and s[left] != " ":
left -= 1
return right - left
C++代码如下:
class Solution {
public:
int lengthOfLastWord(string s) {
int N = s.size();
int left = 0, right = N - 1;
while (right >= 0 && s[right] == ' ') right--;
left = right;
while (left >= 0 && s[left] != ' ') left--;
return right - left;
}
};
单指针
使用一个指针也可以完成上面的操作。代码比较简单,不解释了。
Python版本如下:
class Solution(object):
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
N = len(s)
count = 0
for i in range(N - 1, -1, -1):
if s[i] == " ":
if count == 0:
continue
else:
break
else:
count += 1
return count
C++版本如下:
class Solution {
public:
int lengthOfLastWord(string s) {
int N = s.size();
int count = 0;
for (int i = N - 1; i >= 0; --i) {
if (s[i] == ' ') {
if (count != 0) {
break;
}
} else {
count++;
}
}
return count;
}
};
日期
2017 年 8 月 24 日
2018 年 11 月 24 日 —— 周日开始!一周就过去了~