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:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode() {}
 7  *     ListNode(int val) { this.val = val; }
 8  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 9  * }
10  */
11 class Solution {
12     public int[][] spiralMatrix(int m, int n, ListNode head) {
13         int [][] res = new int[m][n];
14         int r1 = 0;
15         int r2 = m - 1;
16         int c1 = 0;
17         int c2 = n - 1;
18         int ind = 0;
19         while(r1 <= r2 && c1 <= c2){
20             for(int c = c1; c <= c2; c++){
21                 if(head == null){
22                     res[r1][c] = -1;
23                 }else{
24                     res[r1][c] = head.val;
25                     head = head.next;
26                 }
27             }
28 
29             for(int r = r1 + 1; r <= r2; r++){
30                 if(head == null){
31                     res[r][c2] = -1;
32                 }else{
33                     res[r][c2] = head.val;
34                     head = head.next;
35                 }
36             }
37 
38             if(r1 < r2 && c1 < c2){
39                 for(int c = c2 - 1; c >= c1; c--){
40                     if(head == null){
41                         res[r2][c] = -1;
42                     }else{
43                         res[r2][c] = head.val;
44                         head = head.next;
45                     }
46                 }
47 
48                 for(int r = r2 - 1; r > r1; r--){
49                     if(head == null){
50                         res[r][c1] = -1;
51                     }else{
52                         res[r][c1] = head.val;
53                         head = head.next;
54                     }
55                 }
56             }
57 
58             r1++;
59             r2--;
60             c1++;
61             c2--;
62         }
63 
64         return res;
65     }
66 }

类似Spiral Matrix.