###54. Spiral Matrix
题目: https://leetcode.com/problems/spiral-matrix/
难度: Medium
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.