885. Spiral Matrix III
Explanation
To solve this problem, we can simulate the movement in a spiral order by keeping track of the direction we are moving in. We start at the given cell (rStart, cStart) and keep moving in a spiral pattern until we have visited all cells in the grid.
We can define four directions: right, down, left, and up. We start by moving right, then down, then left, then up, and repeat this pattern while adjusting the boundaries of the grid as we move.
We can keep track of the current position and direction, and iterate until we have visited all cells in the grid.
Time Complexity: O(rows * cols) - we need to visit all cells in the grid once.
Space Complexity: O(rows * cols) - the space required to store the result array.
class Solution {
public int[][] spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
int[][] result = new int[rows * cols][2];
int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // right, down, left, up
int direction = 0; // starting with right
int steps = 1; // number of steps to take in current direction
int stepCount = 0; // counter for steps taken
int r = rStart, c = cStart;
int index = 0;
while (index < rows * cols) {
if (r >= 0 && r < rows && c >= 0 && c < cols) {
result[index][0] = r;
result[index][1] = c;
index++;
}
r += directions[direction][0];
c += directions[direction][1];
stepCount++;
if (stepCount == steps) {
direction = (direction + 1) % 4;
if (direction == 0 || direction == 2) {
steps++; // increase steps every two directions
}
stepCount = 0;
}
}
return result;
}
}Code Editor (Testing phase)
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