【LeetCode】面试题29. 顺时针打印数组
题目:
思路:
1、一圈一圈的打印，每次打印一圈，剥洋葱。
2、上下左右四个边界逐次移动。
代码:
Python
class Solution(object):
def printCircle(self, matrix, x1, x2, y1, y2):
# 非二维矩阵时特殊处理
if x1 == x2:
for i in range(y1, y2 + 1):
self.result.append(matrix[x1][i])
elif y1 == y2:
for i in range(x1, x2 +1):
self.result.append(matrix[i][y1])
# 打印当前二维矩阵最外面一圈
else:
for i in range(y1, y2 + 1):
self.result.append(matrix[x1][i])
for i in range(x1+1, x2+1):
self.result.append(matrix[i][y2])
for i in range(y2 - 1,y1-1, -1):
self.result.append(matrix[x2][i])
for i in range(x2 -1, x1, -1):
self.result.append(matrix[i][y1])
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if len(matrix) == 0:
return []
if len(matrix[0]) == 0:
return []
self.result = []
x1 = 0
y1 = 0
x2 = len(matrix) - 1
y2 = len(matrix[0]) - 1
while x1 <= x2 and y1 <= y2:
self.printCircle(matrix, x1, x2, y1, y2)
x1 = x1 + 1
x2 = x2 - 1
y1 = y1 + 1
y2 = y2 -1
return self.result
class Solution(object):
def spiralOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
if len(matrix) == 0 or len(matrix[0]) == 0:
return []
res = []
t = 0
d = len(matrix) - 1
l = 0
r = len(matrix[0]) - 1
while True:
for i in range(l, r + 1, 1):
res.append(matrix[t][i])
t += 1
if t > d: break
for i in range(t, d + 1, 1):
res.append(matrix[i][r])
r -= 1
if r < l: break
for i in range(r, l - 1, -1):
res.append(matrix[d][i])
d -= 1
if d < t: break
for i in range(d, t - 1, -1):
res.append(matrix[i][l])
l += 1
if l > r: break
return res
