1. Traverse Spiral Matrix

The trick of traverse a matrix in spiral order is that we need to have flags to keep the boundary. Let’s check LeetCode problerm 54. Spiral Matrix

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
func spiralOrder(matrix [][]int) []int {
	rows, columns := len(matrix), len(matrix[0])
	length := rows * columns
	ret := make([]int, length)
	up, down, left, right := 0, rows-1, 0, columns-1
	for i := 0; i < length; {
		for j := left; j <= right; j++ {
			ret[i] = matrix[up][j]
			i++
		}
		for j := up + 1; j <= down; j++ {
			ret[i] = matrix[j][right]
			i++
		}
		if up != down {
			for j := right - 1; j >= left; j-- {
				ret[i] = matrix[down][j]
				i++
			}
		}
		if left != right {
			for j := down - 1; j > up; j-- {
				ret[i] = matrix[j][left]
				i++
			}
		}
		up++
		down--
		left++
		right--
	}
	return ret
}

Golang Playbook: https://play.golang.org/p/MTcaLR-vu0g

Similar to the problem above, there’s another one which the start of the spiral is given as part of the input. 885. Spiral Matrix III

  1. Generate Spiral Matrix

Same as spiral matrix traversal, let’s resolve LeetCode problerm 59. Spiral Matrix II

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
func generateMatrix(n int) [][]int {
	ret := make([][]int, n)
	for i := 0; i < n; i++ {
		ret[i] = make([]int, n)
	}
	total := n * n
	up, down, left, right := 0, n-1, 0, n-1
	for i := 1; i <= total; {
		for j := left; j <= right; j++ {
			ret[up][j] = i
			i++
		}
		for j := up + 1; j <= down; j++ {
			ret[j][right] = i
			i++
		}
		if up != down {
			for j := right - 1; j >= left; j-- {
				ret[down][j] = i
				i++
			}
		}
		if left != right {
			for j := down - 1; j > up; j-- {
				ret[j][left] = i
				i++
			}
		}
		up++
		down--
		left++
		right--
	}
	return ret
}

Golang Playbook: https://play.golang.org/p/q-CBvye-xx6

  1. Spiral Traverse with Dynamic Starting Point

The key points to resolve LeetCode problem 885. Spiral Matrix III :
a. After observing the spiral path, you can get the pattern of the step sequence: 1, 1, 2, 2, 3, 3, 4, 4
b. Figure out a way to make a spiral circle n % 4

With those in mind, we can get a solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
func spiralMatrixIII(rows int, cols int, rStart int, cStart int) [][]int {
	ret := [][]int{{rStart, cStart}}
	directions := [4][2]int{
		{0, 1},
		{1, 0},
		{0, -1},
		{-1, 0},
	}
	for i, d := 0, 0; len(ret) < rows*cols; i++ {
		for j := 0; j < i/2+1; j++ {
			rStart += directions[d%4][0]
			cStart += directions[d%4][1]
			if rStart < 0 || cStart < 0 || rStart >= rows || cStart >= cols {
				continue
			}
			ret = append(ret, []int{rStart, cStart})
		}
		d++
	}
	return ret
}

Golang Playbook: https://play.golang.org/p/bPk8zAyAqe2