题目描述

• 递归和迭代都实现了，递归相对比较好理解，也比较好实现
  

思路 & 代码

递归

• 每次都把最外圈的元素跑完
• 一行 or 一列的时候，跑一次结束
• 边界：left > right || low > height 的时候也 return

class Solution {List<Integer> ans = new ArrayList<>();public List<Integer> spiralOrder(int[][] matrix) {// 试试递归recursion(matrix, 0, matrix.length - 1, 0, matrix[0].length - 1);return ans;}// 一次一圈void recursion(int[][] matrix, int low, int height, int left, int right){if(low > height || left > right){return;}// 一行情况if(height == low){for(int i = left; i <= right; i++){ans.add(matrix[low][i]);}return;}// 一列情况else if(left == right){for(int i = low; i <= height; i++){ans.add(matrix[i][left]);}return;}// 其他情况for(int i = left; i <= right; i++){ans.add(matrix[low][i]);}for(int i = low + 1; i <= height; i++){ans.add(matrix[i][right]);}for(int i = right - 1; i >= left; i--){ans.add(matrix[height][i]);}for(int i = height - 1; i > low; i--){ans.add(matrix[i][left]);}recursion(matrix, low + 1, height - 1, left + 1, right - 1);}
}

迭代

• 贴一贴吧，不过看递归的也够了

class Solution {public List<Integer> spiralOrder(int[][] matrix) {// 迭代写法List<Integer> ans = new ArrayList<>();// 一行的情况if(matrix.length == 1){for(int i = 0; i < matrix[0].length; i++){ans.add(matrix[0][i]);}return ans;}if(matrix[0].length == 1){for(int i = 0; i < matrix.length; i++){ans.add(matrix[i][0]);}return ans;}// 碰壁改变方向，更新墙壁值int left = 0, right = matrix[0].length - 1;int low = 1, height = matrix.length - 1;// 右 -》 下，下 -》左，左 -》上，上 => 右边// 右下左上 0，1，2，3int[][] directions = new int[][]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};int forDir = 0;int[] direction = directions[forDir];int x = 0, y = 0;ans.add(matrix[x][y]);boolean flag = true;while(flag){flag = false;    if(forDir == 0){while(y + 1 <= right){ans.add(matrix[x][y + 1]);flag = true;y++;}forDir++;right--;}else if(forDir == 1){while(x + 1 <= height){ans.add(matrix[x + 1][y]);flag = true;x++;}forDir++;height--;}else if(forDir == 2){while(y - 1 >= left){ans.add(matrix[x][y - 1]);flag = true;y--;}forDir++;left++;}else if(forDir == 3){while(x - 1 >= low){ans.add(matrix[x - 1][y]);flag = true;x--;}forDir = 0;low++;}}return ans;}
}

更新版 - 递归

• 注意单列、单行、递归结束几种情况不能漏！

class Solution {List<Integer> ans = new ArrayList<>();public List<Integer> spiralOrder(int[][] matrix) {get(matrix, 0, 0, matrix.length - 1, matrix[0].length - 1);return ans;}void get(int[][] matrix, int xMin, int yMin, int xMax, int yMax) {if(xMin > xMax || yMin > yMax) {return;}if(xMin == xMax) {for(int j = yMin; j <= yMax; j++) {ans.add(matrix[xMin][j]);}return;}if(yMin == yMax) {for(int i = xMin; i <= xMax; i++) {ans.add(matrix[i][yMin]);}return;}int i = xMin, j = yMin;for(; j < yMax; j++) {ans.add(matrix[i][j]);}for(; i < xMax; i++) {ans.add(matrix[i][j]);}for(; j > yMin; j--) {ans.add(matrix[i][j]);}for(; i > xMin; i--) {ans.add(matrix[i][j]);}get(matrix, xMin + 1, yMin + 1, xMax - 1, yMax - 1);} 
}