给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。
示例 1：
输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]示例 2：
输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]提示：
- m == matrix.length
- n == matrix[i].length
- 1 <= m, n <= 10
- -100 <= matrix[i][j] <= 100
解答：
func spiralOrder(matrix [][]int) []int {
var r []int
m := (len(matrix) + 1) / 2
for i := 0; i < m; i++ {
l := len(matrix[0]) - i
k := len(matrix) - i
for j := i; j < l; j++ {
r = append(r, matrix[i][j])
}
if l-i > 0 {
for j := i + 1; j < k; j++ {
r = append(r, matrix[j][l-1])
}
}
if k-i-1 > 0 {
for j := len(matrix[0]) - i - 2; j >= i; j-- {
r = append(r, matrix[k-1][j])
}
}
if l-i-1 > 0 {
for j := k - 2; j > i; j-- {
r = append(r, matrix[j][i])
}
}
}
return r
}官方解答：
1.模拟
func spiralOrder(matrix [][]int) []int {
if len(matrix) == 0 || len(matrix[0]) == 0 {
return []int{}
}
rows, columns := len(matrix), len(matrix[0])
visited := make([][]bool, rows)
for i := 0; i < rows; i++ {
visited[i] = make([]bool, columns)
}
var (
total = rows * columns
order = make([]int, total)
row, column = 0, 0
directions = [][]int{[]int{0, 1}, []int{1, 0}, []int{0, -1}, []int{-1, 0}}
directionIndex = 0
)
for i := 0; i < total; i++ {
order[i] = matrix[row][column]
visited[row][column] = true
nextRow, nextColumn := row+directions[directionIndex][0], column+directions[directionIndex][1]
if nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn] {
directionIndex = (directionIndex + 1) % 4
}
row += directions[directionIndex][0]
column += directions[directionIndex][1]
}
return order
}2.按层模拟
func spiralOrder(matrix [][]int) []int {
if len(matrix) == 0 || len(matrix[0]) == 0 {
return []int{}
}
var (
rows, columns = len(matrix), len(matrix[0])
order = make([]int, rows*columns)
index = 0
left, right, top, bottom = 0, columns - 1, 0, rows - 1
)
for left <= right && top <= bottom {
for column := left; column <= right; column++ {
order[index] = matrix[top][column]
index++
}
for row := top + 1; row <= bottom; row++ {
order[index] = matrix[row][right]
index++
}
if left < right && top < bottom {
for column := right - 1; column > left; column-- {
order[index] = matrix[bottom][column]
index++
}
for row := bottom; row > top; row-- {
order[index] = matrix[row][left]
index++
}
}
left++
right--
top++
bottom--
}
return order
}