0059-leetcode算法实现之螺旋矩阵II-spiralMatrixII-python&golang实现

给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix 。

示例 1：

输入：n = 3
输出：[[1,2,3],[8,9,4],[7,6,5]]
示例 2：

输入：n = 1
输出：[[1]]

提示：

1 <= n <= 20
通过次数132,799提交次数168,447

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/spiral-matrix-ii

python

# 螺旋矩阵II

class Solution:
    def spiralMatrixII(self, n: int) -> [[int]]:
        """
        模拟行为，时间O(n*n), 空间O(n*n)
        思路：
        -left > right, row-top, top++
        -top > bottom, col-right, right--
        -right > left, row-bottom, bottom--
        -bottom > top, col-left, left++
        :param n:
        :return:
        """
        top,bottom,left,right = 0,n-1,0,n-1 # init boundary
        matrix = [[0 for _ in range(n)] for _ in range(n)] # init matrix
        num, target = 1, n*n
        while num <= target:
            for i in range(left, right+1): # left > right, row-top, top++
                matrix[top][i] = num
                num += 1
            top += 1
            for i in range(top, bottom+1): # top > bottom, col-right, right--
                matrix[i][right] = num
                num += 1
            right -= 1
            for i in range(right, left-1, -1): # right > left, row-bottom, bottom--
                matrix[bottom][i] = num
                num += 1
            bottom -= 1
            for i in range(bottom, top-1, -1): # bottom > top, col-left, left++
                matrix[i][left] = num
                num += 1
            left += 1
        return matrix


if __name__ == "__main__":
    n = 3
    test = Solution()
    print(test.spiralMatrixII(n))

golang

package main

import "fmt"

func main() {
	n := 3
	fmt.Println(spiralMatrixII(n))
}

// 模拟行为，螺旋矩阵II
func spiralMatrixII(n int) [][]int {
	matrix := make([][]int, n)
	for i := 0; i < n; i++ {
		matrix[i] = make([]int, n)
	}
	top, bottom, left, right := 0, n-1, 0, n-1
	num, target := 1, n*n
	for num <= target {
		for i := left; i <= right; i++ {
			matrix[top][i] = num
			num++
		}
		top++
		for i := top; i <= bottom; i++ {
			matrix[i][right] = num
			num++
		}
		right--
		for i := right; i >= left; i-- {
			matrix[bottom][i] = num
			num++
		}
		bottom--
		for i := bottom; i >= top; i-- {
			matrix[i][left] = num
			num++
		}
		left++
	}
	return matrix
}