leetcode -- Spiral Matrix -- 思路简单，但很麻烦

https://leetcode.com/problems/spiral-matrix/

思路1 循环矩阵的左上角

这里我的思路就是 以 左上角 i,j = 0,0为初始点，然后每次都从对角线下移，一直到越界或者matrix[i][j]已经在res里面。这里每次都要记录start_i, start_j,即左上角点，以及end_i, end_j右下角点。

另一个要注意的地方就是可能没有spiral，只有一条垂直线或者水平线。

if i == end_i:
if j == end_j:

这两种情况要考虑进去。

class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix: return []
m, n = len(matrix), len(matrix[0])
#if m == 1: return matrix[0]
#if n == 1: return [x[0] for x in matrix]
res = []
i,j = 0, 0
count = 0
while i < m and j < n and matrix[i][j] not in res:
start_i, start_j = i, j
end_i, end_j = m - count - 1, n - count - 1

if i == end_i:
res.extend(matrix[i][start_j:end_j +1])
break
if j == end_j:
for k in xrange(start_i, end_i + 1):
res.append(matrix[k][j])
break

while j<=end_j:
res.append(matrix[i][j])
j += 1
j -= 1
i += 1
while i<=end_i:
res.append(matrix[i][j])
i += 1
i -= 1
j -= 1
while j>=start_j:
res.append(matrix[i][j])
j -= 1
j += 1
i -= 1
while i>start_i:#exclude the mat[i][j]
res.append(matrix[i][j])
i -= 1

i, j = i + 1, j + 1
count += 1
return res

思路2 循环矩阵的up down left right四条边界

参考

/article/4981590.html

有更简单的方法

class Solution:
# @param matrix, a list of lists of integers
# @return a list of integers
def spiralOrder(self, matrix):
if matrix == []: return []
up = 0; left = 0
down = len(matrix)-1
right = len(matrix[0])-1
direct = 0  # 0: go right   1: go down  2: go left  3: go up
res = []
while True:
if direct == 0:
for i in range(left, right+1):
res.append(matrix[up][i])
up += 1
if direct == 1:
for i in range(up, down+1):
res.append(matrix[i][right])
right -= 1
if direct == 2:
for i in range(right, left-1, -1):
res.append(matrix[down][i])
down -= 1
if direct == 3:
for i in range(down, up-1, -1):
res.append(matrix[i][left])
left += 1
if up > down or left > right: return res#注意这句话
direct = (direct+1) % 4#注意这句话