LeetCode:54. 螺旋矩阵

1、题目描述

给定一个包含 m x n 个元素的矩阵(m 行, n 列),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:
[
 [ 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:
[
  [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]

2、题解

2.1、解法一

  原理:递归

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        print("matrix:",matrix)
        if matrix == []:
            return []
        m,n = len(matrix),len(matrix[0])

        ret = []
        # 上侧,右侧
        for i in range(m):
            if i== 0:
                for j in range(n):
                    tmp = matrix[i][j]
                    ret.append(tmp)
            else:
                if m >1 and i != m-1:
                    tmp = matrix[i][n-1]
                    ret.append(tmp)

        print("ret上:",ret)
        # 下侧
        if m >1:
            l = reversed(matrix[m-1])
            print(l)
            for i in l:
                ret.append(i)

        # 左侧
        if m >2 and n>1:
            for i in range(m-2,0,-1):
                ret.append(matrix[i][0])
        print("ret:",ret)

        # 新矩阵
        new = []
        for i in range(m):
            if i != 0 and i != m-1:
                tmp = []
                for j in range(n):
                    if j != 0 and j != n-1:
                        tmp.append(matrix[i][j])
                if tmp != []:
                    new.append(tmp)
        print("new:",new)
        if new != []:
            r = self.spiralOrder(new)
            print("r:",r)
            ret.extend(r)
        return ret

2.2、解法二

  原理: 取首行,去除首行后,对矩阵翻转来创建新的矩阵,再递归直到新矩阵为[],退出并将取到的数据返回

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        # 取首行,去除首行后,对矩阵翻转来创建新的矩阵,
        # 再递归直到新矩阵为[],退出并将取到的数据返回
        ret = []
        if matrix == []:
            return ret
        ret.extend(matrix[0]) # 上侧
        new = [reversed(i) for i in matrix[1:]]
        if new == []:
            return ret
        r = self.spiralOrder([i for i in zip(*new)])
        ret.extend(r)
        return ret

  

posted @ 2018-12-04 17:19  RobotsRising  阅读(443)  评论(0编辑  收藏  举报