54. Spiral Matrix

[Tcdian]

54. Spiral Matrix

给定一个包含 m x n 个元素的矩阵（m 行, n 列），请按照顺时针螺旋顺序，返回矩阵中的所有元素。

Example 1

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]

Example 2

Input:
[
  [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]

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function(matrix) {
    if (matrix.length === 0 || matrix[0].length === 0) {
        return [];
    }
    const direction = [0, 1, 0, -1, 0];
    const visited = new Set();
    const reuslt = [];
    let patrolX = 0;
    let patrolY = -1;
    for (let i = 0; i < 4; i = (i + 1) % 4) {
        patrolX += direction[i];
        patrolY += direction[i + 1]
        while(
            patrolX >= 0
                && patrolX < matrix.length
                && patrolY >= 0
                && patrolY < matrix[0].length
                && !visited.has(`${patrolX}-${patrolY}`)
        ) {
            visited.add(`${patrolX}-${patrolY}`);
            reuslt.push(matrix[patrolX][patrolY]);
            patrolX += direction[i];
            patrolY += direction[i + 1];
        }
        if (visited.size === matrix.length * matrix[0].length) {
            break;
        } else {
            patrolX -= direction[i];
            patrolY -= direction[i + 1];
        }
    }
    return reuslt;
};

• TypeScript Solution

function spiralOrder(matrix: number[][]): number[] {
    if (matrix.length === 0 || matrix[0].length === 0) {
        return [];
    }
    const direction = [0, 1, 0, -1, 0];
    const visited: Set<string> = new Set();
    const reuslt: number[] = [];
    let patrolX = 0;
    let patrolY = -1;
    for (let i = 0; i < 4; i = (i + 1) % 4) {
        patrolX += direction[i];
        patrolY += direction[i + 1]
        while(
            patrolX >= 0
                && patrolX < matrix.length
                && patrolY >= 0
                && patrolY < matrix[0].length
                && !visited.has(`${patrolX}-${patrolY}`)
        ) {
            visited.add(`${patrolX}-${patrolY}`);
            reuslt.push(matrix[patrolX][patrolY]);
            patrolX += direction[i];
            patrolY += direction[i + 1];
        }
        if (visited.size === matrix.length * matrix[0].length) {
            break;
        } else {
            patrolX -= direction[i];
            patrolY -= direction[i + 1];
        }
    }
    return reuslt;
};