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#yyds干货盘点# LeetCode面试题:螺旋矩阵

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1.简述: 给你一个 m 行 n 列的矩阵matrix ,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。 示例 1: 输入:matrix = [[1,2,3],[4,5,6],[7,8,9]] 输出:[1,2,3,6,9,8,7,4,5] 示例 2: 输入:matrix = [

1.简述:

给你一个 m 行 n 列的矩阵 matrix ,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。

 

示例 1:

输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]

输出:[1,2,3,6,9,8,7,4,5]

示例 2:

输入:matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

输出:[1,2,3,4,8,12,11,10,9,5,6,7]

2.代码实现:

class Solution { public List<Integer> spiralOrder(int[][] matrix) { List<Integer> order = new ArrayList<Integer>(); if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { return order; } int rows = matrix.length, columns = matrix[0].length; boolean[][] visited = new boolean[rows][columns]; int total = rows * columns; int row = 0, column = 0; int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; int directionIndex = 0; for (int i = 0; i < total; i++) { order.add(matrix[row][column]); visited[row][column] = true; int nextRow = row + directions[directionIndex][0], nextColumn = column + directions[directionIndex][1]; if (nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn]) { directionIndex = (directionIndex + 1) % 4; } row += directions[directionIndex][0]; column += directions[directionIndex][1]; } return order; }}
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