leetcode-hot100矩阵专题

73.矩阵置零

题目链接

73. 矩阵置零 - 力扣（LeetCode）

解题代码

class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        mapx = []
        mapy = []
        x = len(matrix)
        y = len(matrix[0])
        for i in range(x):
            for j in range(y):
                if matrix[i][j] == 0:
                    mapx.append(i)
                    mapy.append(j)
        for i in mapx:
            for j in range(y):
                matrix[i][j] = 0
        
        for i in mapy:
            for j in range(x):
                matrix[j][i] = 0

54.螺旋矩阵

题目链接

54. 螺旋矩阵 - 力扣（LeetCode）

解题思路

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        if not matrix or not matrix[0]:
            return 
        rows, columns = len(matrix), len(matrix[0])
        visited = [[False] * columns for _ in range(rows)]
        total = rows * columns
        order = [0] * total

        directions = [[0,1],[1,0],[0,-1],[-1,0]]
        row,column = 0,0
        directionIndex = 0
        for i in range(total):
            order[i] = matrix[row][column]
            visited[row][column] = True
            nextRow, nextColumn = row + directions[directionIndex][0],column + directions[directionIndex][1]
            if not (0<=nextRow<rows and 0<=nextColumn<columns and not visited[nextRow][nextColumn]):
                directionIndex = (directionIndex + 1)% 4
            row += directions[directionIndex][0]
            column += directions[directionIndex][1]
        return order