LeetCode 54 Spiral Matrix (模拟蛇形填数)

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本文介绍了一种算法，用于按螺旋顺序遍历二维矩阵的所有元素。通过两种不同的实现方式，展示了如何从左上角开始，顺时针螺旋遍历整个矩阵。

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]

You should return [1,2,3,6,9,8,7,4,5].

题目链接：https://leetcode.com/problems/spiral-matrix/

题目分析：模拟填数过程即可

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> ans = new ArrayList<>();
        int n = matrix.length;
        if (n == 0) {
            return ans;
        }
        int m = matrix[0].length;
        int all = n * m, cnt = 0, i = 0, j = 0;
        boolean[][] vis = new boolean[n + 1][m + 1];
        while (cnt < all) {
            while (j < m && !vis[i][j]) {
                vis[i][j] = true;
                cnt ++;
                ans.add(matrix[i][j ++]);
            }
            j --;
            i ++;
            while (i < n && !vis[i][j]) {
                vis[i][j] = true;
                cnt ++;
                ans.add(matrix[i ++][j]);
            }
            i --;
            j --;
            while (j >= 0 && !vis[i][j]) {
                vis[i][j] = true;
                cnt ++;
                ans.add(matrix[i][j --]);
            }
            j ++;
            i --;
            while (i >= 0 && !vis[i][j]) {
                vis[i][j] = true;
                cnt ++;
                ans.add(matrix[i --][j]);
            }
            i ++;
            j ++;
        }
        return ans;
    }
}

其实不开boolean数组也可以

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> ans = new ArrayList<>();
        int n = matrix.length;
        if (n == 0) {
            return ans;
        }
        int m = matrix[0].length;
        int left = 0, right = m - 1, down = n - 1, up = 0;
        while (true) {
            for (int j = left; j <= right; j ++) {
                ans.add(matrix[up][j]);
            }
            up ++;
            if (left > right || up > down) {
                break;
            }
            for (int i = up; i <= down; i ++) {
                ans.add(matrix[i][right]);
            }
            right --;
            if (left > right || up > down) {
                break;
            }
            for (int j = right; j >= left; j --) {
                ans.add(matrix[down][j]);
            }
            down --;
            if (left > right || up > down) {
                break;
            }
            for (int i = down; i >= up; i --) {
                ans.add(matrix[i][left]);
            }
            left ++;
            if (left > right || up > down) {
                break;
            }
        }
        return ans;
    }
}