LeetCode 2326. Spiral Matrix IV

原题链接在这里：https://leetcode.com/problems/spiral-matrix-iv/description/

题目：

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

题解：

It is the same spriral traversal.
The while condition is r1 <= r2 && c1 <= c2. 

For the last row or last column, we only need to traverse to the right or down, no need to traverse to the left or up.

Check r1 < r2 and c1 < c2 to see if there is only one row or column left.

Time Complexity: O(m * n).

Space: O(1). regardless res.

AC Java:

/**
 * 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];
        int r1 = 0;
        int r2 = m - 1;
        int c1 = 0;
        int c2 = n - 1;
        int ind = 0;
        while(r1 <= r2 && c1 <= c2){
            for(int c = c1; c <= c2; c++){
                if(head == null){
                    res[r1][c] = -1;
                }else{
                    res[r1][c] = head.val;
                    head = head.next;
                }
            }

            for(int r = r1 + 1; r <= r2; r++){
                if(head == null){
                    res[r][c2] = -1;
                }else{
                    res[r][c2] = head.val;
                    head = head.next;
                }
            }

            if(r1 < r2 && c1 < c2){
                for(int c = c2 - 1; c >= c1; c--){
                    if(head == null){
                        res[r2][c] = -1;
                    }else{
                        res[r2][c] = head.val;
                        head = head.next;
                    }
                }

                for(int r = r2 - 1; r > r1; r--){
                    if(head == null){
                        res[r][c1] = -1;
                    }else{
                        res[r][c1] = head.val;
                        head = head.next;
                    }
                }
            }

            r1++;
            r2--;
            c1++;
            c2--;
        }

        return res;
    }
}

类似Spiral Matrix.