LeetCode-in-Go

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

func spiralOrder(matrix [][]int) []int {
	list := make([]int, 0)
	r := 0
	c := 0
	bigR := len(matrix) - 1
	bigC := len(matrix[0]) - 1
	for r <= bigR && c <= bigC {
		for i := c; i <= bigC; i++ {
			list = append(list, matrix[r][i])
		}
		r++
		for i := r; i <= bigR; i++ {
			list = append(list, matrix[i][bigC])
		}
		bigC--
		for i := bigC; i >= c && r <= bigR; i-- {
			list = append(list, matrix[bigR][i])
		}
		bigR--
		for i := bigR; i >= r && c <= bigC; i-- {
			list = append(list, matrix[i][c])
		}
		c++
	}
	return list
}

This site is open source. Improve this page.