Sikoay

将给定字符串按照锯齿形状转换！

由于最后的规则写得太渣所以运行很慢！先贴出来再优化！

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        result = ''
        strlen = len(s)
        if numRows == 1 or strlen == 0:
            return s
        elif numRows == 2:
            list1 = []
            list2 = []
            i = 0
            while i < strlen:
                list1.append(s[i])
                i += 1
                if i >= strlen:
                    break
                list2.append(s[i])
                i += 1
            print list1
            print list2
            list1 = list1 + list2
            return result.join(list1)
        rows = (strlen / (numRows+1))*2
        if strlen % (numRows+1) != 0 :
            rows =  rows + 2
        currentRow = 0
        print rows
        i = 0
        d = dict()
        col = 0
        while currentRow < rows:
            col = 0
            dict2 = dict()
            if currentRow%2==0:
                while col < numRows:
                    if i < strlen:
                        dict2[col] = s[i]
                    else:
                        dict2[col] = ''
                    i += 1
                    col += 1
            else:
                maxi = i+numRows - 2
                curi = maxi - 1
                dict2[0] = ''
                col = 1
                while i < maxi:
                    if curi < strlen:
                        dict2[col] = s[curi]
                    else:
                        dict2[col] = ''
                    i += 1
                    curi -= 1
                    col += 1
                dict2[numRows-1] = ''
            d[currentRow] = dict2
            currentRow += 1
        i = 0
        while i < numRows:
            rowi = 0
            while rowi < rows:
                result += d[rowi][i]
                rowi += 1
            i += 1
        return result

QAQ 弱成狗