115ms python solution

  • 0

    instead of the straight forward way of filling the array with sequence in s, and then concatenating, I took the more difficult way of directly figuring out the exact sequence of characters on each row of the zigzag. It turns out to run faster, as the time for concatenation is saved.

    class Solution:
        # @return a string
        def convert(self, s, nRows):
        	if len(s)<nRows or nRows==1:
        		return s
        	zigSize=nRows*2-2 # number of characters in each zig (down-going and up-going combined).
        	zigNum=len(s)/zigSize # number of such whole zigs needed.
        	hangover=len(s)%zigSize #number of leftover characters in the last incomplete zig
        	secHang=hangover-nRows if hangover > nRows else 0 #number of leftover characters in the last incomplete zig that belong to the up-going arm
        	for i in xrange(nRows):
        		for j in xrange(zigNum):
        			result+=s[zigSize*j+i]+(s[zigSize*(j+1)-i] if (i!=0 and i!=nRows-1) else '')
        		if i < hangover:
        			if i >=nRows-1-secHang and i < nRows-1:
        	return result

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.