Python小技巧：蛇形方阵

输入：数字$m$，$n$
输出：$m$行$n$列的数字蛇形方阵

1. 从方阵最左上角开始，顺时针向内填充。

初始化一个$m$行$n$列的矩阵matrix，所有元素都为0；然后从第一个位置开始，按顺序填充数字，填满了一列或一行后，就转向填写下一列或下一行，最后打印出这个数字蛇形方阵。代码如下：

def print_snake_matrix(m, n):
    matrix = [[0]*n for i in range(m)]
    num = 1
    i, j = 0, 0
    while num <= m*n:
        while j < n and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            j += 1
        j -= 1
        i += 1
        while i < m and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            i += 1
        i -= 1
        j -= 1
        while j >= 0 and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            j -= 1
        j += 1
        i -= 1
        while i >= 0 and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            i -= 1
        i += 1
        j += 1
    for row in matrix:
        for num in row:
            print(num, end='\t')
        print()

执行

print_snake_matrix(4, 6)

输出

1 2 3 4 5 6
16 17 18 19 20 7
15 24 23 22 21 8
14 13 12 11 10 9

c++实现

#include <iostream>
#include <vector>
using namespace std;

void print_snake_matrix(int m, int n) {
    
    
    vector<vector<int>> matrix(m, vector<int>(n, 0));
    int num = 1;
    int i = 0, j = 0;

    while (num <= m * n) {
    
    
        while (j < n && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            j++;
        }
        j--;
        i++;
        while (i < m && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            i++;
        }
        i--;
        j--;
        while (j >= 0 && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            j--;
        }
        j++;
        i--;
        while (i >= 0 && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            i--;
        }
        i++;
        j++;
    }

    for (auto row : matrix) {
    
    
        for (int num : row) {
    
    
            cout << num << "\t";
        }
        cout << endl;
    }
}

2. 从方阵最右上角开始，逆时针向内填充。

由于是从右上角开始填充，所以初始位置为第$0$行第$n-1$列。随后的填充顺序也需要逆时针旋转，以保证填充的数字顺序正确。除此之外，这段代码的实现与前面的代码基本相同，都是采用四个while循环来按顺序填充数字。代码如下：

def print_snake_matrix(m, n):
    matrix = [[0]*n for i in range(m)]
    num = 1
    i, j = 0, n-1
    while num <= m*n:
        while j >= 0 and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            j -= 1
        j += 1
        i += 1
        while i < m and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            i += 1
        i -= 1
        j += 1
        while j < n and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            j += 1
        j -= 1
        i -= 1
        while i >= 0 and matrix[i][j] == 0:
            matrix[i][j] = num
            num += 1
            i -= 1
        i += 1
        j -= 1
    for row in matrix:
        for num in row:
            print(num, end='\t')
        print()

执行

print_snake_matrix(4, 6)

输出

6 5 4 3 2 1
7 20 19 18 17 16
8 21 22 23 24 15
9 10 11 12 13 14

c++实现

void print_snake_matrix(int m, int n) {
    
    
    vector<vector<int>> matrix(m, vector<int>(n, 0));
    int num = 1;
    int i = 0, j = n - 1;

    while (num <= m * n) {
    
    
        while (j >= 0 && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            j--;
        }
        j++;
        i++;
        while (i < m && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            i++;
        }
        i--;
        j++;
        while (j < n && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            j++;
        }
        j--;
        i--;
        while (i >= 0 && matrix[i][j] == 0) {
    
    
            matrix[i][j] = num;
            num++;
            i--;
        }
        i++;
        j--;
    }

    for (auto row : matrix) {
    
    
        for (int num : row) {
    
    
            cout << num << "\t";
        }
        cout << endl;
    }
}