將一個給定字符串根據給定的行數,以從上往下、從左到右進行 Z 字形排列。算法
好比輸入字符串爲 "LEETCODEISHIRING"
行數爲 3 時,排列以下:app
L C I R E T O E S I I G E D H N
以後,你的輸出須要從左往右逐行讀取,產生出一個新的字符串,好比:"LCIRETOESIIGEDHN"
。函數
請你實現這個將字符串進行指定行數變換的函數:spa
string convert(string s, int numRows);
示例 1:code
輸入: s = "LEETCODEISHIRING", numRows = 3 輸出: "LCIRETOESIIGEDHN"
示例 2:blog
輸入: s = "LEETCODEISHIRING", numRows = 4 輸出: "LDREOEIIECIHNTSG" 解釋: L D R E O E I I E C I H N T S G
思路字符串
按照與逐行讀取 Z 字形圖案相同的順序訪問字符串。get
算法string
首先訪問 行 0
中的全部字符,接着訪問 行 1
,而後 行 2
,依此類推...io
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)