leetcode 58. The last word length (Python)

Solution:
This problem is solved with a reverse traversal:

Expect to see this problem in both cases;

1. The last one is the word
2. The last is a space
   for the first case, a direct calculation of the last word in length, then return to;
   but this is also divided into two smaller cases,
   1. Finally there is a word in front of the word, it is determined whether or not the last word in a condition to end the current element pointer is a space;
   2. The last word is not in front of the word, so that the end of the current pointer position of -1;

For the second case, the need to filter out the space, and then calculates; but if the string only spaces, all spaces filtered off, the current pointer position of -1, 0 can be directly returned.

code show as below:

class Solution:
	def lengthOfLastWord(self, s):
		
		right = len(s) - 1

		while right >= 0:
		
			if s[right] != ' ':

				if right  == 0:
					return 1

				count = 0

				while s[right] != ' ' and right >= 0:

					right -= 1
					count += 1					

				return count
			else:
				while s[right] == ' ' and right >= 0:
					right -= 1
				if right == -1:

					return 0
		return 0

Screenshot Results:

The
IF right == 0:
return. 1
are summarized into the first for loop

class Solution:
	def lengthOfLastWord(self, s):
		
		# if len(s) == 1 and s[0] != ' ':

		# 	return 1
		# elif len(s) == 1 and s[0] == ' ':
		# 	return 0

		right = len(s) - 1

		while right >= 0:


		
			if s[right] != ' ':

				
				count = 0

				while s[right] != ' ' and right >= 0:

					right -= 1
					count += 1


				return count
			else:

				while s[right] == ' ' and right >= 0:
					right -= 1
				if right == -1:

					return 0
		return 0

Computing first filtering space, and then the word

class Solution:
	def lengthOfLastWord(self, s):

		end = len(s) - 1
		while (end >= 0 and s[end] == ' '):
			end -= 1
		if end == -1:
			return 0
		start = end
		while start >= 0 and s[start] != ' ':
			start -= 1
		return end - start