[LeetCode][python3]0006. ZigZag Conversion

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N2I -2020.03.15

1. My first try

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        gap=(numRows-1)*2
        if len(s)<2 or gap<=0:
            return s
            
        ans=[]
        for i in range(numRows):
            if i==0 or i==numRows-1:
                ans+=s[i::gap]
                #print(ans)
            else:
                
                dic={}
                index=0
                for item in s[i::gap]:
                    dic[index]=item
                    index+=2
                index=1
                for item in s[(gap-i)::gap]:
                    dic[index]=item
                    index+=2
                for index in range(len(dic)):
                    ans.append(dic[index])        
                #print(ans)
        return ''.join(ans)

Explanation:

Just another slow solution.

2. A better solution

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        #n strings and add them together
        listy = [''] * numRows
        sign, counter = 1, 0
        if numRows == 1:
            return s
        else:
            for i in s:
                listy[counter] += i
                counter = counter + sign
                if counter == numRows - 1:
                    sign = -1
                if counter == 0:
                    sign = 1
            
            finalStr = ''.join(listy)
            return finalStr

Explanation:

The solution use a counter to decide which row the i char will be thrown into. And throw all char in string by sequence. It is more simple and faster.