矩阵模拟问题合集（Leetcode题解-Python语言）

54. 螺旋矩阵（剑指 Offer 29. 顺时针打印矩阵）

class Solution:def spiralOrder(self, matrix: List[List[int]]) -> List[int]:ans = []count = 0m, n = len(matrix), len(matrix[0])length = m * ndirections = [(0, 1), (1, 0), (0, -1), (-1, 0)]x = y = Dir = 0while count < length:ans.append(matrix[x][y])matrix[x][y] = 'a'dx, dy = directions[Dir]if x + dx >= m or y + dy >= n or x + dx < 0 or y + dy < 0 or matrix[x + dx][y + dy] == 'a':Dir = Dir + 1 if Dir < 3 else 0dx, dy = directions[Dir]x += dxy += dycount += 1return ans

首先考虑到，元素前进的方向只有四个，对应方向的 x 与 y 如何变化是确定的。所以用一个direction 数组保存四个方向 [(0, 1), (1, 0), (0, -1), (-1, 0)]。另外要注意防止跑出界，每次准备跑出界时就转向。对于已经取过数的格子，把它置 ‘a’ 即可，下一格为 ‘a’ 就转向。

59. 螺旋矩阵 II

class Solution:def generateMatrix(self, n: int) -> List[List[int]]:matrix = [[0] * n for _ in range(n)]directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]x = y = Dir = 0for i in range(1, n**2 + 1):matrix[x][y] = idx, dy = directions[Dir]if x + dx >= n or y + dy >= n or x + dx < 0 or y + dy < 0 or matrix[x + dx][y + dy] != 0:Dir = Dir + 1 if Dir < 3 else 0dx, dy = directions[Dir]x += dxy += dyreturn matrix

与上一题类似，区别在于本题不是给定矩阵求遍历序列，而是给定一个数字 n 然后返回一个 n**2 大小的矩阵。

885. 螺旋矩阵 III

class Solution:def spiralMatrixIII(self, rows: int, cols: int, rStart: int, cStart: int) -> List[List[int]]:ans= []directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]Left, Right, Upper, Bottom = cStart - 1, cStart + 1, rStart - 1, rStart + 1  # 四个方向的边界x, y, num, Dir = rStart, cStart, 1, 0  # (x, y)为当前节点，num为当前查找的数字，Dir为当前的方向while num <= rows * cols:if x >= 0 and x < rows and y >= 0 and y < cols:  # (x, y)在矩阵中ans.append([x, y])num += 1if Dir == 0 and y == Right:  # 向右到右边界Dir += 1    # 调转方向向下Right += 1  # 右边界右移elif Dir == 1 and x == Bottom:  # 向下到底边界Dir += 1Bottom += 1  # 底边界下移elif Dir == 2 and y == Left:  # 向左到左边界Dir += 1Left -= 1  # 左边界左移elif Dir == 3 and x == Upper:  # 向上到上边界Dir = 0Upper -= 1  # 上边界上移dx, dy = directions[Dir]x += dxy += dyreturn ans