用 Go 剑指 Offer 29. 顺时针打印矩阵

发布时间 2023-04-07 17:08:09作者: slowlydance2me

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

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/spiral-matrix
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

 

消消乐法:顺时针走,读取一行就删一行

func spiralOrder(matrix [][]int) []int {
    if len(matrix) == 0 || len(matrix[0]) == 0{
        return []int{}
    }
    res := make([]int, 0)
    up, left, down, right := 0, 0, len(matrix)-1, len(matrix[0])-1
    for {
        for i := left; i <= right; i++ {
            res = append(res, matrix[up][i])
        }
        up++
        if up > down {
            break
        }
        for i := up; i <= down; i++ {
            res = append(res, matrix[i][right])
        }
        right--
        if right < left {
            break
        }
        for i := right; i >= left; i-- {
            res = append(res, matrix[down][i])
        }
        down--
        if down < up {
            break
        }
        for i := down; i >= up; i-- {
            res = append(res, matrix[i][left])
        }
        left++
        if left > right {
            break
        }
    }
    return res
}