DAY23：leetcode #59 Spiral Matrix II

最新推荐文章于 2023-12-14 23:03:25 发布

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最新推荐文章于 2023-12-14 23:03:25 发布

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分类专栏：算法文章标签： leetcode python

本文链接：https://blog.youkuaiyun.com/zcy19941015/article/details/53087013

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本文介绍了一个生成螺旋矩阵的算法，该算法接收一个整数n作为输入，并返回一个n×n的矩阵，矩阵中的元素从1到n²按螺旋顺序填充。文章通过递归的方式实现了这一过程，详细解释了如何一层一层地构建螺旋矩阵。

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Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

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class Solution(object):
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        self.result = []
        for i in range(n):
            self.result.append([0]*n)
        self.handleMatrix(0,1,n)
        return self.result

    def handleMatrix(self, start,startnum,n):
        if startnum > n**2:
            return
        for j in range(start, n-start):
            self.result[start][j] = startnum
            startnum += 1
        for i in range(start + 1, n-start):
            self.result[i][n - start - 1] = startnum
            startnum += 1
        for j in range(start, n - start - 1)[::-1]:
            self.result[n-start-1][j] = startnum
            startnum += 1
        for i in range(start + 1, n - start - 1)[::-1]:
            self.result[i][start] = startnum
            startnum += 1
        self.handleMatrix(start + 1,startnum,n)

思路基本与54类似，不再赘述