You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
Example:
[[0,1,0,0],
[1,1,1,0],
[0,1,0,0],
[1,1,0,0]]
Answer: 16
Explanation: The perimeter is the 16 yellow stripes in the image below:
我的解法：就是数数而已
class Solution(object):
def islandPerimeter(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
# for each element in island, calculate stripes and sum them
row = len(grid)
col = len(grid[0])
ans = 0
for i in range(0, row):
for j in range(0, col):
if grid[i][j] == 1:
if i==0 or grid[i-1][j] == 0:
ans += 1
if i==row-1 or grid[i+1][j] == 0:
ans += 1
if j==0 or grid[i][j-1] == 0:
ans += 1
if j==col-1 or grid[i][j+1]==0:
ans += 1
return ans更简单的做法，统计为1的矩形个数x，以及上下和左右相邻的边数y，结果为4x-2y
class Solution(object):
def islandPerimeter(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
# for each element in island, calculate stripes and sum them
row = len(grid)
col = len(grid[0])
ans = 0
for i in range(0, row):
for j in range(0, col):
if grid[i][j] == 1:
ans += 4
if i>0 and grid[i-1][j] == 1:
ans -= 2
if j>0 and grid[i][j-1] == 1:
ans -= 2
return ans还有一种DFS版本，
class Solution(object):
def islandPerimeter(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
# for each element in island, calculate stripes and sum them
row = len(grid)
col = len(grid[0])
self.ans = 0
def dfs(grid,i,j):
if i<0 or i>=row or j<0 or j>=col:
return
if grid[i][j] == 0: return
if grid[i][j] == 1:
grid[i][j] = -1
if i==0 or grid[i-1][j] == 0:
self.ans += 1
if i==row-1 or grid[i+1][j] == 0:
self.ans += 1
if j==0 or grid[i][j-1] == 0:
self.ans += 1
if j==col-1 or grid[i][j+1]==0:
self.ans += 1
dfs(grid, i-1, j)
dfs(grid, i+1, j)
dfs(grid, i, j-1)
dfs(grid, i, j+1)
for i in range(0, row):
for j in range(0, col):
if grid[i][j] == 1:
dfs(grid, i, j)
return self.ansDFS的代码结构：
dfs(xxx):
process(xxx.val)
dfs(xxx.left)
dfs(xxx.right)也就和树的深度优先，先序遍历类似！
