Question

输入是一个N x N矩阵，必须循环移位（顺时针或逆时针）一个元素。

示例输入（尺寸3）：

1 2 3
4 5 6
7 8 9

输出

2 3 6
1 5 9
4 7 8

示例输入（尺寸4）：

10 20 5  5
22 33 4  10
2  6  8  9
55 11 66 7

输出

22 10 20 5 
2  6  33 5 
55 8  4  10
11 66 7  9

我需要知道如何解决这个问题。

Answer 1

要进行轮换，请按顺序阅读元素：
我 - ＆gt; 0到n-1和j = 0
j - ＆gt; 0到n-1，i = n-1
我 - ＆gt; n-1到0和j = n-1
j - ＆gt; n-1到0且i = 0

import collections
def shiftMatrix(matrix, layers, num,rotate_direction):
    for count in range(layers):
        temp1 = []
        i = count
        for j in range(count, num - count):
            temp1.append(matrix[i][j])
        j = num - 1 - count
        for i in range(1 + count, num - count -1):
            temp1.append(matrix[i][j])
        i = num - 1 - count
        for j in range(num - 1 - count, -1 + count, -1):
            temp1.append(matrix[i][j])
        j = count
        for i in range(num - 1 - count - 1, 0 + count, -1):
            temp1.append(matrix[i][j])
        temp = collections.deque(temp1)
        temp.rotate(rotate_direction)
        temp1 = list(temp)
        counter = 0
        i = count
        for j in range(count, num - count):
            matrix[i][j] = temp1[counter]
            counter += 1
        j = num - 1 - count
        for i in range(1 + count, num - count -1):
            matrix[i][j] = temp1[counter]
            counter += 1
        i = num - 1 - count
        for j in range(num - 1 - count, -1 + count, -1):
            matrix[i][j] = temp1[counter]
            counter += 1
        j = count
        for i in range(num - 1 - count - 1, 0 + count, -1):
            matrix[i][j] = temp1[counter]
            counter += 1
        return matrix


if __name__ == '__main__':
    num = int(raw_input())
    op_type = raw_input()
    matrix = []
    for i in range(num):
        temp = []
        input_string = raw_input().split()
        for element in input_string:
            temp.append(int(element))
        matrix.append(temp)
    if(num >= 3):
        layers = num - 3 + 1
    else:
        layers = 1
    if(op_type == "CW"):
        matrix_result = shiftMatrix(matrix,layers,num,1);
    else:
        matrix_result = shiftMatrix(matrix,layers,num,-1);
    print(" ".join((" ".join(map(str,element))) for element in matrix_result))

说明： 尺寸N的矩阵的层数是N-3。对于每一层，元素已经以顺时针方式收集。然后转换成一个顺时针或逆时针旋转的队列。

顺时针移动矩阵

1 个答案: