Given an m x n matrix, return all elements of the matrix in spiral order.
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
- logic in straight forward, lots of book keeping
- in a while loop, adjust boundry while pushing values in one direction
var spiralOrder = function (matrix) {
const result = [];
if (!matrix.length) {
return result;
}
let startRow = 0;
let endRow = matrix.length - 1;
let startCol = 0;
let endCol = matrix[0].length - 1;
let dir = 0;
while (startRow <= endRow && startCol <= endCol) {
switch (dir) {
case 0: //RIGHT
for (let col = startCol; col <= endCol; col++) {
result.push(matrix[startRow][col]);
}
startRow++;
break;
case 1: //DOWN
for (let row = startRow; row <= endRow; row++) {
result.push(matrix[row][endCol]);
}
endCol--;
break;
case 2: //LEFT
for (let col = endCol; col >= startCol; col--) {
result.push(matrix[endRow][col]);
}
endRow--;
break;
case 3: //UP
for (let row = endRow; row >= startRow; row--) {
result.push(matrix[row][startCol]);
}
startCol++;
break;
}
dir = (dir + 1) % 4;
}
return result;
};
Time and Space Complexity
