将一个给定字符串根据给定的行数,以从上往下、从左到右进行 Z 字形排列。
比如输入字符串为 "LEETCODEISHIRING"
行数为 3 时,排列如下:
L C I R E T O E S I I G E D H N
之后,你的输出需要从左往右逐行读取,产生出一个新的字符串,比如:"LCIRETOESIIGEDHN"
。
请你实现这个将字符串进行指定行数变换的函数:
string convert(string s, int numRows);
示例 1:
输入: s = "LEETCODEISHIRING", numRows = 3 输出: "LCIRETOESIIGEDHN"
示例 2:
输入: s = "LEETCODEISHIRING", numRows = 4 输出: "LDREOEIIECIHNTSG" 解释: L D R E O E I I E C I H N T S G
思路
按照与逐行读取 Z 字形图案相同的顺序访问字符串。
算法
首先访问 行 0
中的所有字符,接着访问 行 1
,然后 行 2
,依此类推...
解法一:
class Solution(object): def convert(self, s, numRows): """ :type s: str :type numRows: int :rtype: str """ if numRows <= 1: return s n = len(s) ans = [] step = 2 * numRows - 2 for i in range(numRows): one = i two = -i while one < n or two < n: if 0 <= two < n and one != two and i != numRows - 1: ans.append(s[two]) if one < n: ans.append(s[one]) one += step two += step return "".join(ans)
解法二:
class Solution: def convert(self, s, numRows): """ :type s: str :type numRows: int :rtype: str """ # no need to convert if numRows == 1: return(s) zlist = [] sc = "" n = numRows # create null list while n: zlist.append([]) n = n - 1 j = 0 for a in s: if j == 0: # direction change coverse = False zlist[j].append(a) if j + 1 < numRows: if coverse: j = j - 1 else: j = j + 1 else: j = j - 1 # direction change coverse = True # get the converted string for z in zlist: for t in z: sc = sc + t return(sc)