代码随想录算法训练营Day 63| 图论 part03 | 417.太平洋大西洋水流问题、827.最大人工岛、127. 单词接龙
文章目录
- [代码随想录算法训练营Day 63| 图论 part03 | 417.太平洋大西洋水流问题、827.最大人工岛、127. 单词接龙](#代码随想录算法训练营Day 63| 图论 part03 | 417.太平洋大西洋水流问题、827.最大人工岛、127. 单词接龙)
- 17.太平洋大西洋水流问题
- 827.最大人工岛
-
- [一、DFS 用全局变量得到area](#一、DFS 用全局变量得到area)
- [二、DFS 用局部变量](#二、DFS 用局部变量)
- 三、BFS
- [127. 单词接龙](#127. 单词接龙)
17.太平洋大西洋水流问题
一、DFS
python
class Solution(object):
def pacificAtlantic(self, heights):
"""
:type heights: List[List[int]]
:rtype: List[List[int]]
"""
m,n=len(heights),len(heights[0])
dirs = [(-1,0),(0,1),(1,0),(0,-1)]
pacific=[[0]*n for _ in range(m)]
atlantic=[[0]*n for _ in range(m)]
result=[]
# DFS
def dfs(x,y,ocean):
ocean[x][y]=1
for d in dirs:
nextx,nexty=x+d[0],y+d[1]
if 0 <= nextx < m and 0 <= nexty < n and heights[nextx][nexty] >=heights[x][y] and ocean[nextx][nexty]==0:
dfs(nextx,nexty,ocean)
for i in range(m):
dfs(i,0,pacific)
dfs(i,n-1,atlantic)
for j in range(n):
dfs(0,j,pacific)
dfs(m-1,j,atlantic)
for i in range(m):
for j in range(n):
if pacific[i][j]==1 and atlantic[i][j]==1:
result.append([i,j])
return result
二、BFS
python
class Solution(object):
def pacificAtlantic(self, heights):
"""
:type heights: List[List[int]]
:rtype: List[List[int]]
"""
m,n=len(heights),len(heights[0])
dirs = [(-1,0),(0,1),(1,0),(0,-1)]
pacific=[[0]*n for _ in range(m)]
atlantic=[[0]*n for _ in range(m)]
result=[]
# BFS
def bfs(x,y,ocean):
q=collections.deque()
q.append((x,y))
ocean[x][y]=1
while q:
x,y = q.popleft()
for d in dirs:
nextx,nexty=x+d[0],y+d[1]
if 0 <= nextx < m and 0 <= nexty < n and heights[nextx][nexty] >=heights[x][y] and ocean[nextx][nexty]==0:
ocean[nextx][nexty]=1
q.append((nextx,nexty))
for i in range(m):
bfs(i,0,pacific)
bfs(i,n-1,atlantic)
for j in range(n):
bfs(0,j,pacific)
bfs(m-1,j,atlantic)
for i in range(m):
for j in range(n):
if pacific[i][j]==1 and atlantic[i][j]==1:
result.append([i,j])
return result三、本题总结
用两个visited来表示
827.最大人工岛
一、DFS 用全局变量得到area
python
class Solution(object):
def largestIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
'''
总体思路
利用 DFS 计算出各个岛屿的面积，并标记每个 1（陆地格子）属于哪个岛。
遍历每个 0，统计其上下左右四个相邻格子所属岛屿的编号，去重后，累加这些岛的面积，更新答案的最大值。
'''
m,n = len(grid),len(grid[0])
dirs = [(-1,0),(0,1),(1,0),(0,-1)]
area = collections.defaultdict(int) # 用于储存岛屿面积
def dfs(x,y,island_num): # 输入岛屿编号
grid[x][y]=island_num
area[island_num] += 1 # 更新岛屿面积
for d in dirs:
nextx,nexty=x+d[0],y+d[1]
if 0 <= nextx < m and 0 <= nexty < n and grid[nextx][nexty]==1:
grid[nextx][nexty]=island_num
dfs(nextx,nexty,island_num)
island_num = 1
for i in range(m):
for j in range(n):
if grid[i][j]==1: # 遇到新岛屿
island_num += 1 # 岛屿编号从2开始
dfs(i,j,island_num)
ans=0
for i in range(m):
for j in range(n):
s=set() # 去重
if grid[i][j]==0:
for d in dirs:
nexti,nextj=i+d[0],j+d[1]
if 0 <= nexti < m and 0 <= nextj < n and grid[nexti][nextj]!=0:
s.add(grid[nexti][nextj])
ans = max(ans,1+sum(area[idx] for idx in s))
return ans if ans else n*n # 如果最后 ans 仍然为 0，说明所有格子都是 1，返回 n^2二、DFS 用局部变量
python
class Solution(object):
def largestIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
'''
总体思路
利用 DFS 计算出各个岛屿的面积，并标记每个 1（陆地格子）属于哪个岛。
遍历每个 0，统计其上下左右四个相邻格子所属岛屿的编号，去重后，累加这些岛的面积，更新答案的最大值。
'''
m,n = len(grid),len(grid[0])
dirs = [(-1,0),(0,1),(1,0),(0,-1)]
area = collections.defaultdict(int) # 用于储存岛屿面积
def dfs(x,y,island_num): # 输入岛屿编号
grid[x][y]=island_num
size=1
# area[island_num] += 1 # 更新岛屿面积
for d in dirs:
nextx,nexty=x+d[0],y+d[1]
if 0 <= nextx < m and 0 <= nexty < n and grid[nextx][nexty]==1:
grid[nextx][nexty]=island_num
size += dfs(nextx,nexty,island_num)
return size # 得到岛屿的面积
island_num = 1
for i in range(m):
for j in range(n):
if grid[i][j]==1: # 遇到新岛屿
island_num += 1 # 岛屿编号从2开始
area[island_num]=dfs(i,j,island_num)
ans=0
for i in range(m):
for j in range(n):
s=set() # 去重
if grid[i][j]==0:
for d in dirs:
nexti,nextj=i+d[0],j+d[1]
if 0 <= nexti < m and 0 <= nextj < n and grid[nexti][nextj]!=0:
s.add(grid[nexti][nextj])
ans = max(ans,1+sum(area[idx] for idx in s))
return ans if ans else n*n # 如果最后 ans 仍然为 0，说明所有格子都是 1，返回 n^2三、BFS
python
class Solution(object):
def largestIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
'''
总体思路
利用 DFS 计算出各个岛屿的面积，并标记每个 1（陆地格子）属于哪个岛。
遍历每个 0，统计其上下左右四个相邻格子所属岛屿的编号，去重后，累加这些岛的面积，更新答案的最大值。
'''
# BFS
def bfs(x,y,island_num): # 输入岛屿编号
grid[x][y]=island_num
size=1
# area[island_num] += 1 # 更新岛屿面积
q=collections.deque()
q.append((x,y))
while q:
x,y=q.popleft()
for d in dirs:
nextx,nexty=x+d[0],y+d[1]
if 0 <= nextx < m and 0 <= nexty < n and grid[nextx][nexty]==1:
grid[nextx][nexty]=island_num
q.append((nextx,nexty))
size += 1
return size
island_num = 1
for i in range(m):
for j in range(n):
if grid[i][j]==1: # 遇到新岛屿
island_num += 1 # 岛屿编号从2开始
# dfs(i,j,island_num) # 法1
area[island_num]=bfs(i,j,island_num)
ans=0
for i in range(m):
for j in range(n):
s=set() # 去重
if grid[i][j]==0:
for d in dirs:
nexti,nextj=i+d[0],j+d[1]
if 0 <= nexti < m and 0 <= nextj < n and grid[nexti][nextj]!=0:
s.add(grid[nexti][nextj])
ans = max(ans,1+sum(area[idx] for idx in s))
return ans if ans else n*n # 如果最后 ans 仍然为 0，说明所有格子都是 1，返回 n^2127. 单词接龙
一、BFS
python
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
"""
:type beginWord: str
:type endWord: str
:type wordList: List[str]
:rtype: int
"""
wordset = set(wordList)
if len(wordList)==0 or endWord not in wordset :
return 0
q = collections.deque()
q.append(beginWord)
visited=set(beginWord)
step=1
while q:
level = len(q)
for l in range(level):
word = q.popleft()
word_list = list(word)
for i in range(len(word_list)):
origin_char=word_list[i]
for j in range(26):
word_list[i] = chr(ord('a')+j)
new_word = ''.join(word_list)
if new_word in wordset:
if new_word == endWord:
return step+1
if new_word not in visited:
q.append(new_word)
visited.add(new_word)
word_list[i]=origin_char
step +=1
return 0