58. Length of Last Word

题目：

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

.

链接： http://leetcode.com/problems/length-of-last-word/

一刷，还是有粗心错误。从后向前遍历，初始化in_word为False，之后如果不为空则count += 1，如果为空则继续往前直到查找到第一个非空的字母。再向前查找空字符，找到或者循环结束时返回。

class Solution(object):
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
in_word = False
count = 0
for rev_idx in range(len(s), 0, -1):
val = s[rev_idx - 1]
if not in_word and val == ' ':
continue
elif in_word and val == ' ':
return count
elif val != ' ':
in_word = True
count += 1
else:
return count if in_word else 0

参考别人代码简化：in_word和count是否为0是一个状态，可省略一个。

class Solution(object):
def lengthOfLastWord(self, s):
"""
:type s: str
:rtype: int
"""
count = 0
for rev_idx in range(len(s), 0, -1):
val = s[rev_idx - 1]
if val == ' ':
if not count:
continue
else:
break
else:
count += 1
return count

2/12/2017, Java

错误：

1. 空格和''没有区分好

2. 第6行的判断条件应该是start < 0，不是start == 0

3. 最后返回时候注意没有＋1，需要多用例子测试

public class Solution {
public int lengthOfLastWord(String s) {
int start = s.length() - 1;

while (start >= 0 && s.charAt(start) == ' ') start--;
if (start < 0) return 0;
int end = start;
while (end >= 0 && s.charAt(end) != ' ') end--;
return start - end;
}
}