. Length of Last Word

Topic:

Given A string s consists of upper/lower-case alphabets and empty space characters ‘ ‘ , return the length of LA St Word in the string.

If the last word does not exist, return 0.

Note:a word is defined as A character sequence consists for non-space characters only.

For example,
Given S = "Hello World" ,
Return 5 .

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

A brush, or there is a careless mistake. From backward forward, initialize In_word to False, then count + = 1 if not empty, and continue forward until you find the first non-empty letter if it is empty. The null character is then searched forward, either found or returned at the end of the loop.

classsolution (object):defLengthoflastword (self, s):""": Type S:str:rtype:int"""In_word=False Count=0 forRev_idxinchRange (len (s), 0, 1): Val= S[rev_idx-1]            if  notIn_word andval = =' ':                Continue            elifIn_word andval = =' ':                returnCountelifVal! =' ': In_word=True Count+ = 1Else:            returnCountifIn_wordElse0

Refer to others code simplification: whether In_word and Count 0 is a state, omit one.

classsolution (object):defLengthoflastword (self, s):""": Type S:str:rtype:int"""Count=0 forRev_idxinchRange (len (s), 0, 1): Val= S[rev_idx-1]            ifval = =' ':                if  notCount:Continue                Else:                     Break            Else: Count+ = 1returnCount

. Length of Last Word