Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
Example:
Input: 3Output:
Walk and Turn Algorithm to Fill the Matrix in Sprial Clock-wise Order
We start at the top-left corner where we fill number 1, then the initial direction is RIGHT, then we keep walking until we hit the border or the cell has been filled already. Then we turn right, repeatedly doing this until we have finished the matrix.
The special case is the 1×1 matrix, we can just immediately return [1] without walking. The following is the Java implementation of the Clock-wise spiral matrix. In Java, we use Arrays.fill to initialize a one-dimension array. We can use a for loop to initialize a two dimensional array using Arrays.fill.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | class Solution { public int[][] generateMatrix(int n) { int[][] res = new int[n][n]; for (int[] x: res) { Arrays.fill(x, -1); } if (n <= 1) { res[0][0] = 1; return res; } int d[][] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; int x = 0, r = 0, c = 0, num = 1; int total = n * n; res[0][0] = 1; while (num < total) { int nr = r + d[x][0]; int nc = c + d[x][1]; if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) { x = (x + 1) % 4; } else { r = nr; c = nc; res[r][c] = ++ num; } } return res; } } |
class Solution {
public int[][] generateMatrix(int n) {
int[][] res = new int[n][n];
for (int[] x: res) {
Arrays.fill(x, -1);
}
if (n <= 1) {
res[0][0] = 1;
return res;
}
int d[][] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int x = 0, r = 0, c = 0, num = 1;
int total = n * n;
res[0][0] = 1;
while (num < total) {
int nr = r + d[x][0];
int nc = c + d[x][1];
if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) {
x = (x + 1) % 4;
} else {
r = nr;
c = nc;
res[r][c] = ++ num;
}
}
return res;
}
}Similarly, the following is the C++ version of the Spiral square matrix.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | class Solution { public: vector<vector>int>> generateMatrix(int n) { vector<vector<int>> res(n, vector<int>(n, -1)); if (n <= 1) { return vector<vector<int>>(1, {1}); } int d[][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; int x = 0, r = 0, c = 0, num = 1; int total = n * n; res[0][0] = 1; while (num < total) { int nr = r + d[x][0]; int nc = c + d[x][1]; if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) { x = (x + 1) % 4; } else { r = nr; c = nc; res[r][c] = ++ num; } } return res; } }; |
class Solution {
public:
vector<vector>int>> generateMatrix(int n) {
vector<vector<int>> res(n, vector<int>(n, -1));
if (n <= 1) {
return vector<vector<int>>(1, {1});
}
int d[][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int x = 0, r = 0, c = 0, num = 1;
int total = n * n;
res[0][0] = 1;
while (num < total) {
int nr = r + d[x][0];
int nc = c + d[x][1];
if (nr >= n || nc >= n || nr < 0 || nc < 0 || res[nr][nc] != -1) {
x = (x + 1) % 4;
} else {
r = nr;
c = nc;
res[r][c] = ++ num;
}
}
return res;
}
};Syntaxally much the same, both implementations run O(N^2) time and space complexity.
See also: Construct a Spiral Matrix using Simulation Algorithm
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