python 实现二维列表转置

python 二维列表转置

def transpose(self, matrix):
    new_matrix = []
    for i in range(len(matrix[0])):
      matrix1 = []
      for j in range(len(matrix)):
        matrix1.append(matrix[j][i])
      new_matrix.append(matrix1)
    return new_matrix

python 二维列表逆时针转置

def transpose(self, matrix):
    new_matrix = []
    for i in range(len(matrix[0])):
      matrix1 = []
      for j in range(len(matrix)):
        matrix1.append(matrix[j][i])
      new_matrix.append(matrix1)
    return new_matrix[::-1]

例子：

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字，例如，如果输入如下4 X 4矩阵： 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 则依次打印出数字1,2,3,4,8,12,16,15,14,13,9,5,6,7,11,10.

# -*- coding:utf-8 -*-
class Solution:
  # matrix类型为二维列表，需要返回列表
  # matrix应该是列表组成的列表
  # 去掉首行，然后逆时针转置
  def printMatrix(self, matrix):
    # write code here
    result = []
    while matrix:
      result.extend(matrix.pop(0))
      if not matrix:
        break
      matrix = self.transpose(matrix)
    return result
  # 转置
  def transpose(self, matrix):
    new_matrix = []
    for i in range(len(matrix[0])):
      matrix1 = []
      for j in range(len(matrix)):
        matrix1.append(matrix[j][i])
      new_matrix.append(matrix1)
    return new_matrix[::-1]