9.7 练手 腾讯50题 54螺旋矩阵

Leecode 54. Spiral Matrix

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]

感觉有点像在写个游戏，和以前写的扫雷判断条件有点像，比较简单。

代码如下：

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        x = len(matrix)
        if(x == 0):
            return matrix
        if(x == 1):
            return matrix[0]
        y = len(matrix[0])
        turnR = True
        turnL = False
        Down = False
        Up = False
        sumNum = x*y
        curx, cury = 0, 0
        WallR = y-1
        WallL = 0
        WallBottom = x-1
        WallTop = 1
        output = [matrix[curx][cury]]
        sumNum -= 1
        #print(finalX,finalY)
        while(sumNum>0):
            if(cury==WallR and turnR):
                turnR = False
                Down = True
                WallR -= 1
            if(cury==WallL and turnL):
                turnL = False
                Up = True
                WallL += 1
            if(curx==WallTop and Up):
                Up = False
                turnR =True
                WallTop += 1
            if(curx==WallBottom and Down):
                Down = False
                turnL = True
                WallBottom -= 1
            if(turnR):
                cury += 1
            if(turnL):
                cury -= 1
            if(Down):
                curx += 1
            if(Up):
                curx -= 1
            sumNum -= 1
            output += [matrix[curx][cury]]
            
        return output

时间 40%+ 复杂度为O（n），感觉时间不是很快啊

看了discuss 里面的代码，发现一整块数组一起搬回更快，一个一个还是慢了，这边用x in Karray 用得很棒，学习一下编一下。

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if matrix == []:
            return []
        if len(matrix) == 1:
            return matrix[0]
        if len(matrix[0]) == 1:
            return [x[0] for x in matrix]
        top = 0
        bottom = len(matrix)
        left = 0
        right = len(matrix[0])
        res = []
        while top<bottom and right>left:
            res += matrix[top][left:right]
            top += 1
            res += [x[right-1] for x in matrix[top:bottom]] 
            right -= 1
            if top < bottom:
                res += matrix[bottom-1][right-1:left:-1]
            bottom -= 1
            if left < right:
                res += [x[left] for x in matrix[bottom:top-1:-1]]             
            left += 1
        return res