0054. 螺旋矩阵

[utterances-bot]

0054. 螺旋矩阵 | 算法通关手册

54. 螺旋矩阵 # 标签：数组 难度：中等 题目大意 # 描述：给定一个 m * n 大小的二维矩阵 matrix。 要求：按照顺时针旋转的顺序，返回矩阵中的所有元素。

https://algo.itcharge.cn/Solutions/0001-0099/spiral-matrix/

[Huang131]

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        res = []
        while matrix:
            res += matrix.pop(0)
            if not matrix:
                break
            matrix = self.turnMatrix(matrix)
        return res

    def turnMatrix(self, matrix: List[List[int]]) -> List[int]:
        rows = len(matrix)
        cols = len(matrix[0])

        newmat = []
        for i in range(cols):
            newrow = []
            for j in range(rows):
                newrow.append(matrix[j][i])
            newmat.append(newrow)
        newmat.reverse()
        return newmat

[s1acr]

class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
ans = []
x, y = 0, 0
dx, dy = 0, 1
nx, ny = 0, 0
m = len(matrix)
n = len(matrix[0])
up, down, left, right = 0, m-1, 0, n-1

    while True:
        if nx < up or nx > down or ny < left or ny > right:
            if nx < up:
                left += 1
                if left > right:
                    break
            elif nx > down:
                right -= 1
                if right < left:
                    break
            elif ny < left:
                down -= 1
                if down < up:
                    break
            elif ny > right:
                up += 1
                if up > down:
                    break
            dx, dy = dy, -dx
            x, y = x + dx, y + dy
        else:
            x, y = nx, ny
        ans.append(matrix[x][y])
        nx, ny = x+dx, y+dy
    return ans