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
螺旋矩陣 IV。
給你兩個整數：m 和 n ，表示矩陣的維數。另給你一個整數連結串列的頭節點 head 。
請你生成一個大小為 m x n 的螺旋矩陣，矩陣包含連結串列中的所有整數。連結串列中的整數從矩陣 左上角 開始、順時針 按 螺旋 順序填充。如果還存在剩餘的空格，則用 -1 填充。
返回生成的矩陣。
思路
還是按照類似54題那樣的遍歷方式，無非是把 node.val 放入座標裡。
複雜度
時間O(mn)
空間O(mn) - output matrix
程式碼
Java實現
/**
* 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;
}
}