# 994. Rotting Oranges
# In a given grid, each cell can have one of three values:
# 0 --- empty cell;
# 1 --- fresh orange;
# 2 --- rotten orange.
# Every minute, adjacent fresh to rotten
# minimum minutes till no fresh orange.
class Solution:
def orangesRotting(self, G: List[List[int]]) -> int:
M, N, S, E, c = len(G), len(G[0]), [], sum(G,[]).count(1), 0
R = [(i,j) for i,j in itertools.product(range(M),range(N)) if G[i][j] == 2]
# product(list1, list2) 依次取出list1中的每1个元素,与list2中的每1个元素,组成元组,
# 然后,将所有的元组组成一个列表,
while E != 0:
for [i,j] in R:
for a,b in (i-1,j),(i,j+1),(i+1,j),(i,j-1):
if 0 <= a < M and 0 <= b < N and G[a][b] == 1:
G[a][b], E, _ = 2, E - 1, S.append([a,b])
if len(S) == 0: return -1
R, S, c = S, [], c + 1
return c
# 127. Word Ladder
# Only one letter can be changed at a time.
# beginWord and endWord; dictionary's word list
# length of shortest transformation begin to end;
class Solution(object):
def ladderLength(self, beginWord, endWord, wordList):
wordList = set(wordList) #字典
queue = collections.deque([[beginWord, 1]]) #
# collections.deque
while queue:
word, length = queue.popleft() #
if word == endWord:
return length
for i in range(len(word)):
for c in 'abcdefghijklmnopqrstuvwxyz':
next_word = word[:i] + c + word[i+1:]
if next_word in wordList:
wordList.remove(next_word)
queue.append([next_word, length + 1])
return 0
# collections 是 python 内建的一个集合模块,里面封装了许多集合类;
# 其中队列相关的集合只有一个:deque;
# deque 是双边队列(double-ended queue),具有队列和栈的性质;
# 在 list 的基础上增加了移动、旋转和增删等;
# d = collections.deque([])
# d.append('a') # 在最右边添加一个元素,此时 d=deque('a')
# d.appendleft('b') # 在最左边添加一个元素,此时 d=deque(['b', 'a'])
# d.extend(['c','d']) # 在最右边添加所有元素,此时 d=deque(['b', 'a', 'c', 'd'])
# d.extendleft(['e','f']) # 在最左边添加所有元素,此时 d=deque(['f', 'e', 'b', 'a', 'c', 'd'])
# d.pop() # 将最右边的元素取出,返回 'd',此时 d=deque(['f', 'e', 'b', 'a', 'c'])
# d.popleft() # 将最左边的元素取出,返回 'f',此时 d=deque(['e', 'b', 'a', 'c'])
# d.rotate(-2) # 向左旋转两个位置(正数则向右旋转),此时 d=deque(['a', 'c', 'e', 'b'])
# d.count('a') # 队列中'a'的个数,返回 1
# d.remove('c') # 从队列中将'c'删除,此时 d=deque(['a', 'e', 'b'])
# d.reverse() # 将队列倒序,此时 d=deque(['b', 'e', 'a'])
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