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54. Spiral Matrix

难度: Medium

刷题内容

原题连接

内容描述

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]

解题方案

思路 1 - 时间复杂度: O(m * n)- 空间复杂度: O(1)******

参考别人的代码，一开始觉得很有递归性，根据奇偶不同来写，递归太难写。

然后想到了loop，再想，可能有更优trick，事实证明并没有。

用四个变量来控制边界，然后因为方向总是：→↓←↑ 左右下上

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if matrix == [] : return []
        res = []
        maxUp = maxLeft = 0
        maxDown = len(matrix) - 1
        maxRight = len(matrix[0]) - 1 
        direction = 0 # 0 go right, 1 go down, 2 go left, 3 up
        while True:
            if direction == 0: #go right
                for i in range(maxLeft, maxRight+1):
                    res.append(matrix[maxUp][i])
                maxUp += 1
            elif direction == 1: # go down
                for i in range(maxUp, maxDown+1):
                    res.append(matrix[i][maxRight])
                maxRight -= 1
            elif direction == 2: # go left
                for i in reversed(range(maxLeft, maxRight+1)):
                    res.append(matrix[maxDown][i])
                maxDown -= 1
            else: #go up
                for i in reversed(range(maxUp, maxDown+1)):
                    res.append(matrix[i][maxLeft])
                maxLeft +=1
            if maxUp > maxDown or maxLeft > maxRight:
                return res
            direction = (direction + 1 ) % 4

以上的写法非常精妙，看看我自己用同样的思路写的||||

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

        left = 0
        up = 0
        down = len(matrix) - 1
        right = len(matrix[0]) - 1

        # 0 -> right, 1 -> down, 2-> left, 3 -> up
        direction = 0

        # start location
        x, y = 0, 0
        res = []

        while True:
            if left > right or up > down:
                return res

            if direction == 0:
                while y <= right:
                    res.append(matrix[up][y])
                    y += 1
                up += 1
                x = up
                direction = 1
                continue

            if direction == 1:
                while x <= down:
                    res.append(matrix[x][right])
                    x += 1
                right -= 1
                y = right
                direction = 2
                continue

            if direction == 2:
                while y >= left:
                    res.append(matrix[down][y])
                    y -= 1
                down -= 1
                x = down
                direction = 3
                continue

            if direction == 3:
                while x >= up:
                    res.append(matrix[x][left])
                    x -= 1
                left += 1
                y = left
                direction = 0
                continue

明显别人的代码写的更精妙，因为这里两个boundary都很明确，所以用for in range就能很好的解决问题了.

最后放一个无敌一行，怕你看完不想看上面的代码了

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        return matrix and list(matrix.pop(0)) + self.spiralOrder(list(zip(*matrix))[::-1])

oh, my god!