## Array - 54. Spiral Matrix

Updated on with 0 views and 0 comments

54.Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

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]
``````

go：

``````func spiralOrder(matrix [][]int) []int {
var  res []int
if matrix == nil || len(matrix) == 0 {
return res
}
printGrid(matrix, 0, len(matrix) -1, 0, len(matrix[0]) - 1, &res)
return res
}

func printGrid(grid [][]int, top, bottom, left, right int, res *[]int) {
// 递归出口
if top > bottom || left > right {
return
}

if top == bottom {  // 1. 只有一列的情况
for i := left; i <= right; i++ {
*res = append(*res, grid[top][i])
}
} else if left == right {  // 2. 只有一行的情况
for i := top; i <= bottom; i++ {
*res = append(*res, grid[i][left])
}
} else {  // 3.正常绕圈打印
// 从左往右打印
for i := left; i <= right - 1; i++ {
*res = append(*res, grid[top][i])
}

// 从上往下打印
for i := top; i <= bottom - 1; i++ {
*res = append(*res, grid[i][right])
}

// 从右往左打印
for i := right; i >= left + 1; i-- {
*res = append(*res, grid[bottom][i])
}

// 从下往上打印
for i := bottom; i >= top + 1; i-- {
*res = append(*res, grid[i][left])
}
}

printGrid(grid, top+1, bottom-1, left+1, right-1, res)
}
``````