LeetCode in TypeScript

54. Spiral Matrix

Medium

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]

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100

Solution

function spiralOrder(matrix: number[][]): number[] {
    const result: number[] = []
    let r = 0,
        c = 0
    let bigR = matrix.length - 1
    let bigC = matrix[0].length - 1
    while (r <= bigR && c <= bigC) {
        for (let i = c; i <= bigC; i++) {
            result.push(matrix[r][i])
        }
        r++
        for (let i = r; i <= bigR; i++) {
            result.push(matrix[i][bigC])
        }
        bigC--
        for (let i = bigC; i >= c && r <= bigR; i--) {
            result.push(matrix[bigR][i])
        }
        bigR--
        for (let i = bigR; i >= r && c <= bigC; i--) {
            result.push(matrix[i][c])
        }
        c++
    }
    return result
}

export { spiralOrder }

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