Spiral Matrix-蒲公英云

Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

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

You should return [1,2,3,6,9,8,7,4,5].

剑指offer的题目．很有意思，但是非常考验细节的题目,主体细节在circle这个函数，每次走四个方向．从左朝右，从上朝下，从右朝左，从下朝上．和剑指offer的思路不一样．我没有加过多判断．而是要求每次按行按列走都要走完，走完之后，修改index，将该行或者该列删除．复杂度Ｏ(m*n),代码如下：

class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if not matrix or not matrix[0]:
            return []
        res = []
        m = len(matrix)
        n = len(matrix[0])
        xBegin = 0
        xEnd = m - 1
        yBegin = 0
        yEnd = n - 1
        self.circle(res,matrix, xBegin, xEnd, yBegin, yEnd)
        return res
    def circle(self, res, matrix, xBegin, xEnd, yBegin, yEnd):
        if xBegin <= xEnd and  yBegin <= yEnd:
            if yBegin <= yEnd and xBegin <= xEnd: #这一行可以不需要，去掉会加速很多
                for i in xrange(yBegin, yEnd+1):
                    res.append(matrix[xBegin][i])
                xBegin += 1
            if xBegin <= xEnd and yBegin <= yEnd:
                for i in xrange(xBegin, xEnd+1):
                    res.append(matrix[i][yEnd])
                yEnd -= 1
            if yBegin <= yEnd and xBegin <= xEnd:
                for j in xrange(yEnd, yBegin-1, -1):
                    res.append(matrix[xEnd][j])
                xEnd -= 1
            if xBegin <= xEnd and yBegin <= yEnd:
                for j in xrange(xEnd, xBegin-1, -1):
                    res.append(matrix[j][yBegin])
                yBegin += 1
            self.circle(res,matrix, xBegin, xEnd, yBegin, yEnd)