[LeetCode] 2326. Spiral Matrix IV

You are given two integers m and n, which represent the dimensions of a matrix.

You are also given the head of a linked list of integers.

Generate an m x n matrix that contains the integers in the linked list presented in spiral order (clockwise), starting from the top-left of the matrix. If there are remaining empty spaces, fill them with -1.

Return the generated matrix.

Example 1:

Input: m = 3, n = 5, head = [3,0,2,6,8,1,7,9,4,2,5,5,0]
Output: [[3,0,2,6,8],[5,0,-1,-1,1],[5,2,4,9,7]]
Explanation: The diagram above shows how the values are printed in the matrix.
Note that the remaining spaces in the matrix are filled with -1.

Example 2:

Input: m = 1, n = 4, head = [0,1,2]
Output: [[0,1,2,-1]]
Explanation: The diagram above shows how the values are printed from left to right in the matrix.
The last space in the matrix is set to -1.

Constraints:
1 <= m, n <= 105
1 <= m * n <= 105
The number of nodes in the list is in the range [1, m * n].
0 <= Node.val <= 1000

螺旋矩阵 IV。

给你两个整数：m 和 n ，表示矩阵的维数。

另给你一个整数链表的头节点 head 。

请你生成一个大小为 m x n 的螺旋矩阵，矩阵包含链表中的所有整数。链表中的整数从矩阵 左上角 开始、顺时针 按 螺旋 顺序填充。如果还存在剩余的空格，则用 -1 填充。

返回生成的矩阵。

思路

还是按照类似54题那样的遍历方式，无非是把 node.val 放入坐标里。

复杂度

时间O(mn)
空间O(mn) - output matrix

代码

Java实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int[][] spiralMatrix(int m, int n, ListNode head) {
        int[][] res = new int[m][n];
        // initial the matrix with -1
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                res[i][j] = -1;
            }
        }
 
        int top = 0;
        int bottom = m - 1;
        int left = 0;
        int right = n - 1;
        ListNode cur = head;
        while (left <= right && top <= bottom) {
            // left to right
			for (int i = left; i <= right; i++) {
				if (cur != null) {
					res[top][i] = cur.val;
					cur = cur.next;
				} else {
					break;
				}
			}
			top++;
 
            // right to bottom
            for (int i = top; i <= bottom; i++) {
				if (cur != null) {
					res[i][right] = cur.val;
					cur = cur.next;
				} else {
					break;
				}
			}
			right--;
 
            // right to left
            if (top <= bottom) {
				for (int i = right; i >= left; i--) {
					if (cur != null) {
						res[bottom][i] = cur.val;
						cur = cur.next;
					} else {
						break;
					}
				}
			}
			bottom--;
 
            // bottom to top
            if (left <= right) {
				for (int i = bottom; i >= top; i--) {
					if (cur != null) {
						res[i][left] = cur.val;
						cur = cur.next;
					} else {
						break;
					}
				}
			}
			left++;
 
        }
        return res;
    }
}