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]