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Formatted question description: https://leetcode.ca/all/2326.html

# 2326. Spiral Matrix IV

• Difficulty: Medium.
• Related Topics: Array, Linked List, Matrix, Simulation.
• Similar Questions: Spiral Matrix, Spiral Matrix II, Spiral Matrix III.

## Problem

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

## Solution (Java, C++, Python)

• /**
* 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 {
private enum Direction {
RIGHT,
DOWN,
LEFT,
UP
}

public int[][] spiralMatrix(int m, int n, ListNode head) {
int[][] arr = new int[m][n];
int i = 0;
int j = -1;
Direction direction = Direction.RIGHT;
// Boundaries
// ++ after Left to right Horizontal traversed
int a = 0;
// -- after Down to Up vertical traversed
int b = n - 1;
// -- after Right to Left horizontal teversed
int c = m - 1;
// ++ after Down to Up vertical traversed
int d = 0;
for (int k = 0; k < m * n; ++k) {
int val = -1;
}
switch (direction) {
case RIGHT:
++j;
if (j == b) {
direction = Direction.DOWN;
++a;
}
break;
case DOWN:
++i;
if (i == c) {
direction = Direction.LEFT;
}
break;
case LEFT:
--j;
if (j == d) {
--c;
direction = Direction.UP;
}
break;
case UP:
default:
--i;
if (i == a) {
--b;
++d;
direction = Direction.RIGHT;
}
break;
}
arr[i][j] = val;
}
return arr;
}
}

############

/**
* 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[][] ans = new int[m][n];
for (int[] row : ans) {
Arrays.fill(row, -1);
}
int i = 0, j = 0, p = 0;
int[][] dirs = { {0, 1}, {1, 0}, {0, -1}, {-1, 0}};
while (true) {
break;
}
while (true) {
int x = i + dirs[p][0], y = j + dirs[p][1];
if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0) {
p = (p + 1) % 4;
} else {
i = x;
j = y;
break;
}
}
}
return ans;
}
}

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]:
ans = [[-1] * n for _ in range(m)]
i = j = p = 0
dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]]
while 1:
break
while 1:
x, y = i + dirs[p][0], j + dirs[p][1]
if x < 0 or y < 0 or x >= m or y >= n or ~ans[x][y]:
p = (p + 1) % 4
else:
i, j = x, y
break
return ans

############

# 2326. Spiral Matrix IV
# https://leetcode.com/problems/spiral-matrix-iv/

# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def spiralMatrix(self, rows: int, cols: int, head: Optional[ListNode]) -> List[List[int]]:
matrix = [[-1] * cols for _ in range(rows)]

rowStart, rowEnd = 0, rows - 1
colStart, colEnd = 0, cols - 1

# traverse right
i = colStart
while head and i <= colEnd:
i += 1
rowStart += 1

# traverse down
i = rowStart
while head and i <= rowEnd:
i += 1
colEnd -= 1

# traverse left
i = colEnd
while head and i >= colStart:
i -= 1
rowEnd -= 1

# traverse up
i = rowEnd
while head and i >= rowStart:
i -= 1
colStart += 1

return matrix


• /**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
vector<vector<int>> spiralMatrix(int m, int n, ListNode* head) {
vector<vector<int>> ans(m, vector<int>(n, -1));
int i = 0, j = 0, p = 0;
vector<vector<int>> dirs = { {0, 1}, {1, 0}, {0, -1}, {-1, 0} };
while (1) {
while (1) {
int x = i + dirs[p][0], y = j + dirs[p][1];
if (x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0)
p = (p + 1) % 4;
else {
i = x, j = y;
break;
}
}
}
return ans;
}
};

• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
func spiralMatrix(m int, n int, head *ListNode) [][]int {
ans := make([][]int, m)
for i := range ans {
ans[i] = make([]int, n)
for j := range ans[i] {
ans[i][j] = -1
}
}
i, j, p := 0, 0, 0
dirs := [][]int{ {0, 1}, {1, 0}, {0, -1}, {-1, 0} }
for {
break
}
for {
x, y := i+dirs[p][0], j+dirs[p][1]
if x < 0 || y < 0 || x >= m || y >= n || ans[x][y] >= 0 {
p = (p + 1) % 4
} else {
i, j = x, y
break
}
}
}
return ans
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function spiralMatrix(m: number, n: number, head: ListNode | null): number[][] {
const dirs = [
[0, 1],
[1, 0],
[0, -1],
[-1, 0],
];
let ans = Array.from({ length: m }, v => new Array(n).fill(-1));
let i = 0,
j = 0,
k = 0;
let x = i + dirs[k][0];
let y = j + dirs[k][1];
if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || ans[x][y] != -1) {
k = (k + 1) % 4;
}
i = i + dirs[k][0];
j = j + dirs[k][1];
}
return ans;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).