问题描述 LeetCode代码，使用了do while语句来减少判断分支
class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
cells_num = 0
cells_sum = 0
to_check = set()
directions = [(-1,0),(1,0),(0,1),(0,-1)]
flag = 0
for i in range(1, len(grid)-1):
for j in range(1, len(grid[0])-1):
if grid[i][j] == 1:
grid[i][j] = 0
cells_num += 1
ix = i
iy = j
while True:
for dx, dy in directions:
if 0 <= ix+dx < len(grid) and 0 <= iy+dy < len(grid[0]):
if grid[ix+dx][iy+dy] == 1:
grid[ix+dx][iy+dy] = 0
to_check.add((ix+dx, iy+dy))
if ix + dx == 0 or ix + dx == len(grid) - 1 or iy + dy == 0 or iy + dy == len(grid[0]) - 1:
flag = 1
else:
cells_num += 1
if not to_check:
if flag:
flag = 0
else:
cells_sum += cells_num
cells_num = 0
break
ix, iy = to_check.pop()
return cells_sumLeetcode运行结果，速度比上一题有所提升
pycharm代码
class Solution:
def numEnclaves(self, grid):
cells_num = 0
cells_sum = 0
to_check = set()
directions = [(-1,0),(1,0),(0,1),(0,-1)]
flag = 0
for i in range(1, len(grid)-1):
for j in range(1, len(grid[0])-1):
if grid[i][j] == 1:
grid[i][j] = 0
cells_num += 1
ix = i
iy = j
while True:
for dx, dy in directions:
if 0 <= ix+dx < len(grid) and 0 <= iy+dy < len(grid[0]):
if grid[ix+dx][iy+dy] == 1:
grid[ix+dx][iy+dy] = 0
to_check.add((ix+dx, iy+dy))
if ix + dx == 0 or ix + dx == len(grid) - 1 or iy + dy == 0 or iy + dy == len(grid[0]) - 1:
flag = 1
else:
cells_num += 1
if not to_check:
if flag:
flag = 0
else:
cells_sum += cells_num
cells_num = 0
break
ix, iy = to_check.pop()
return cells_sum
grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
print(Solution().numEnclaves(grid))
grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
print(Solution().numEnclaves(grid))
grid = [[0, 0, 0, 1, 1, 1, 0, 1, 0, 0], [1, 1, 0, 0, 0, 1, 0, 1, 1, 1], [0, 0, 0, 1, 1, 1, 0, 1, 0, 0],
[0, 1, 1, 0, 0, 0, 1, 0, 1, 0], [0, 1, 1, 1, 1, 1, 0, 0, 1, 0], [0, 0, 1, 0, 1, 1, 1, 1, 0, 1],
[0, 1, 1, 0, 0, 0, 1, 1, 1, 1], [0, 0, 1, 0, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 0, 0, 0, 1]]
print(Solution().numEnclaves(grid))
