[剑指offer][JAVA]面试题第[29]题[顺时针打印矩阵][数组]

【问题描述】[中等]

输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。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]限制：0 <= matrix.length <= 100
0 <= matrix[i].length <= 100

【解答思路】

1. 模拟

在这里插入图片描述
时间复杂度：O(NM) 空间复杂度：O(NM)

class Solution {public int[] spiralOrder(int[][] matrix) {if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {return new int[0];}int rows = matrix.length, columns = matrix[0].length;boolean[][] visited = new boolean[rows][columns];int total = rows * columns;int[] order = new int[total];int row = 0, column = 0;int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};int directionIndex = 0;//total二维数组的所有个数for (int i = 0; i < total; i++) {order[i] = 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;}
}

2. 按层模拟（圆圆圈圈圆圆）

解释一
在这里插入图片描述

时间复杂度：O(NM) 空间复杂度：O(NM)

class Solution {public int[] spiralOrder(int[][] matrix) {if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {return new int[0];}int rows = matrix.length, columns = matrix[0].length;int[] order = new int[rows * columns];int index = 0;int left = 0, right = columns - 1, top = 0, bottom = rows - 1;while (left <= right && top <= bottom) {for (int column = left; column <= right; column++) {order[index++] = matrix[top][column];}for (int row = top + 1; row <= bottom; row++) {order[index++] = matrix[row][right];}//判断必不可少 if (left < right && top < bottom) {for (int column = right - 1; column > left; column--) {order[index++] = matrix[bottom][column];}for (int row = bottom; row > top; row--) {order[index++] = matrix[row][left];}}left++;right--;top++;bottom--;}return order;}
}

解释二
在这里插入图片描述
时间复杂度：O(NM) 空间复杂度：O(NM)

class Solution {public int[] spiralOrder(int[][] matrix) {if(matrix.length == 0) return new int[0];int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0;int[] res = new int[(r + 1) * (b + 1)];while(true) {for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right.if(++t > b) break;for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom.if(l > --r) break;for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left.if(t > --b) break;for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top.if(++l > r) break;}return res;}
}

【总结】

1.++i 和 i++

++在前下自加后运算；++在后先运算后自加

2.画图有助于理解，定点切分

3.细节学习

3.1 while break结合

while(true) {       if(条件) break;}

3.2 二维矩阵遍历初始化

int rows = matrix.length, columns = matrix[0].length;boolean[][] visited = new boolean[rows][columns];int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};