leetcode--python--之字形打印矩阵

如上图所示的矩阵打印：
方法： 先设置2个指针，移动时a指针向下，b指针向右，当a走到最下面的时候向右移动，当b中移动到最右边的时候，向下移动，知道a==b

def z_print_mtrix(mtrix):
    a_x, a_y, b_x, b_y = 0, 0, 0, 0
    x_length = len(mtrix) - 1
    y_length = len(mtrix[0]) - 1
    reverse_flag = True
    while b_x != x_length + 1: 
        corner_to_crner(reverse_flag, a_x, a_y, b_x, b_y, mtrix)
        if a_x != x_length:
            a_x += 1
        else:
            a_y += 1
        if b_y != y_length:
            b_y += 1
        else:
            b_x += 1

        reverse_flag = not reverse_flag
def corner_to_crner(reverse_flag, a_x, a_y, b_x, b_y, mtrix):#用来遍历a，b所组成的对角线
    if not reverse_flag:
        while a_x != b_x - 1:
            print(mtrix[a_x][a_y], end=' ')
            a_x -= 1
            a_y += 1
    else:
        while b_x != a_x + 1:
            print(mtrix[b_x][b_y], end=' ')
            b_x += 1
            b_y -= 1
li = [[1,2,3,4],
      [5,6,7,8],
      [9,10,11,14]]
z_print_mtrix(li)

如果题目让输出一个m行n列的按之字形递增的矩阵，可以稍微修改下

def z_print_mtrix(mtrix):
    a_x, a_y, b_x, b_y = 0, 0, 0, 0
    x_length = len(mtrix) - 1
    y_length = len(mtrix[0]) - 1
    reverse_flag = True
    m = 1
    while b_x != x_length + 1: 
        m = corner_to_crner(reverse_flag, a_x, a_y, b_x, b_y, mtrix, m)
        if a_x != x_length:
            a_x += 1
        else:
            a_y += 1
        if b_y != y_length:
            b_y += 1
        else:
            b_x += 1
#         m += 1
        reverse_flag = not reverse_flag
def corner_to_crner(reverse_flag, a_x, a_y, b_x, b_y, mtrix, m):#用来遍历a，b所组成的对角线
    if not reverse_flag:
        while a_x != b_x - 1:
            mtrix[a_x][a_y] = m
            a_x -= 1
            a_y += 1
            m+=1
    else:
        while b_x != a_x + 1:
            mtrix[b_x][b_y] = m
            b_x += 1
            b_y -= 1
            m+=1
    return m
mtrix = [[0 for _ in range(4)] for _ in range(4)]
z_print_mtrix(mtrix)

最后

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