Easy, Msc

Question

返回一個(gè)m x n的矩陣的螺旋序列

For example,
矩陣:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
的螺旋序列是[1,2,3,6,9,8,7,4,5].

Solution

螺旋序列的直觀表示如下圖

Screen Shot 2017-08-14 at 8.55.29 PM.jpg

直觀答案如下

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        ret = []
        if len(matrix) == 0: return ret
        
        m, n = len(matrix), len(matrix[0])
        row, col = 0, -1
        while True:
            for i in xrange(n):
                col += 1
                ret.append(matrix[row][col])
            m -= 1
            if m == 0: break
                
            for i in xrange(m):
                row += 1
                ret.append(matrix[row][col])
            n -= 1
            if n == 0: break
                
            for i in xrange(n):
                col -= 1
                ret.append(matrix[row][col])
            m -= 1
            if m == 0: break
                
            for i in xrange(m):
                row -= 1
                ret.append(matrix[row][col])
            n -= 1
            if n == 0: break
                
        return ret       

更加簡(jiǎn)潔的寫(xiě)法

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        ret = []
        if len(matrix) == 0:
            return ret
        while matrix:
            ret += matrix.pop(0)
            if matrix and matrix[0]:
                for row in matrix:
                    ret.append(row.pop())
            if matrix:
                ret += matrix.pop()[::-1]
            if matrix and matrix[0]:
                for row in matrix[::-1]:
                    ret.append(row.pop(0))
        return ret

最后獻(xiàn)上StefanPochmann 大神的神解法, 真正讓矩陣旋轉(zhuǎn)起來(lái)。

def spiralOrder(self, matrix):
    return matrix and list(matrix.pop(0)) + self.spiralOrder(zip(*matrix)[::-1])