标签：bsp obj += ast type integer cas while strip()
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
Java:
public int lengthOfLastWord(String s) {
String use = s.trim();
int count = 0;
for (int i = use.length() - 1; i >= 0; i--) {
if (use.charAt(i) != ‘ ‘) count++;
else break;
}
return count;
}
Java:
public int lengthOfLastWord(String s) {
return s.trim().length()-s.trim().lastIndexOf(" ")-1;
}
Python:
class Solution(object):
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
return len(s.strip().split(‘ ‘)[-1])
Python:
class Solution:
# @param s, a string
# @return an integer
def lengthOfLastWord(self, s):
length = 0
for i in reversed(s):
if i == ‘ ‘:
if length:
break
else:
length += 1
return length
C++:
class Solution {
public:
int lengthOfLastWord(string s) {
int len = 0, tail = s.length() - 1;
while (tail >= 0 && s[tail] == ‘ ‘) tail--;
while (tail >= 0 && s[tail] != ‘ ‘) {
len++;
tail--;
}
return len;
}
};
[LeetCode] 58. Length of Last Word 求末尾单词的长度
标签：bsp obj += ast type integer cas while strip()
原文地址：https://www.cnblogs.com/lightwindy/p/9602354.html
