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SpiralMatrix.java
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SpiralMatrix.java
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// Link: https://leetcode.com/problems/spiral-matrix/
/*
* Recursion
* 0ms, 100%
*
* Layer by layer from outer to inner.
*
* For example,
*
* 1 2 3
* 4 5 6
* 7 8 9
*
* We use four for-loop to traverse each layer. E.g., for the first layer, the first for-loop for 1, 2;
* the second for-loop for 3, 6; the third for-loop for 9, 8 and the last for-loop for 7, 4.
*/
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> ans = new ArrayList<>();
spiral(0, matrix, matrix.length, matrix[0].length, ans);
return ans;
}
private void spiral(int layer, int[][] matrix, int m, int n, List<Integer> ans) {
// Base case: three cases
if (m == 0 || n == 0) {
// No more row or column left
return;
}
if (m == 1) {
// Just one row (the middile row) left, including the case where only one cell (i.e., the center cell) left
for (int i = layer; i < layer + n; ++i) {
ans.add(matrix[layer][i]);
}
return;
}
if (n == 1) {
// Just one column (the middle column) left
for (int i = layer; i < layer + m; ++i) {
ans.add(matrix[i][layer]);
}
return;
}
// Traverse four edges of the current layer in spiral order
// Up
for (int i = layer; i < layer + n - 1; ++i) {
ans.add(matrix[layer][i]);
}
// Right
for (int i = layer; i < layer + m - 1; ++i) {
ans.add(matrix[i][layer + n - 1]);
}
// Down
for (int i = layer + n - 1; i >= layer + 1; --i) {
ans.add(matrix[layer + m - 1][i]);
}
// Left
for (int i = layer + m - 1; i >= layer + 1; --i) {
ans.add(matrix[i][layer]);
}
// Deal with the next layer
spiral(layer + 1, matrix, m - 2, n - 2, ans);
}
}
/*
* Time complexity: O(m*n), m is the number of rows and n is the number of columns
*
* Space complexity: O(min(m, n)) for the recursion stack
*
* Notes:
*
*/