332:重新岸炮行程
机场字典：{起飞机场：[到达机场的列表]}
去重：到达机场列表，i>0时，当前机场和上一个机场相等，continue
1 class Solution:
2 def findItinerary(self, tickets: List[List[str]]) -> List[str]:
3 if(not tickets): return tickets
4 targets={}
5 for i in tickets:
6 if(i[0] not in targets.keys()): targets[i[0]]=[]
7 targets[i[0]].append(i[1])
8 for i in targets: targets[i].sort()
9 re=['JFK']
10 self.backtracking(len(tickets),targets,re,"JFK")
11 return re
12 def backtracking(self,lens,targets,re,tic):
13 if(len(re)==lens+1): return True
14 if(tic not in targets or not targets[tic]): return False
15 for i,dest in enumerate(targets[tic]):
16 if(i>0 and dest==targets[tic][i-1]): continue
17 targets[tic].pop(i)
18 re.append(dest)
19 if(self.backtracking(lens,targets,re,dest)): return True
20 targets[tic].insert(i, dest)
21 re.pop()
22 return FalsefindItinerary
51：N皇后
for遍历行
递归列
同一列没有Q，45° 135°对角线没有Q，可以放
1 import copy
2 class Solution:
3 def solveNQueens(self, n: int) -> List[List[str]]:
4 if(not n): return []
5 #预置棋盘，所有位置均可以放
6 path=[]
7 for i in range(n):
8 path.append(copy.deepcopy(['.']*n))
9 re=[]
10 self.backtracking(n,re,path,0)
11 return re
12 def backtracking(self,n,re,path,startID):
13 #一次遍历完成
14 if(startID==n):
15 re.append(self.get_result(path))
16 return
17 for i in range(n):
18 #是否可以放Q
19 if(self.isValid(startID,i,n,path)):
20 path[startID][i]='Q'
21 self.backtracking(n,re,path,startID+1)
22 #回溯棋盘
23 path[startID][i]='.'
24
25 def isValid(self,x,y,n,path):
26 #一列有Q
27 for i in range(n):
28 if('Q' == path[i][y]): return False
29 i=x-1
30 j=y-1
31 #45°对角线
32 while i>=0 and j>=0:
33 if(path[i][j]=='Q'): return False
34 i-=1
35 j-=1
36 i=x-1
37 j=y+1
38 #135°对角线
39 while i>=0 and j<n:
40 if(path[i][j]=='Q'): return False
41 i-=1
42 j+=1
43 return True
44 def get_result(self,path):
45 re=[]
46 for i in path:
47 s=''
48 for j in i:
49 s+=j
50 re.append(s)
51 return resolveNQueens
37:解数独
(⊙﹏⊙)双层for循环中再加一层for循环，然后居然就过了
怎么说
1 class Solution:
2 def solveSudoku(self, board: List[List[str]]) -> None:
3 """
4 Do not return anything, modify board in-place instead.
5 """
6 for i in range(len(board)):
7 for j in range(len(board[0])):
8 if(board[i][j]!='.'): continue
9 for k in range(1,10):
10 if(self.isValid(i,j,str(k),board)):
11 board[i][j]=str(k)
12 if self.solveSudoku(board): return True
13 board[i][j]='.'
14 return False
15 return True
16 def isValid(self,x,y,k,board):
17 #行
18 if k in board[x]: return False
19 #列
20 for i in range(9):
21 if board[i][y]==k: return False
22 xi=x-x%3
23 yj=y-y%3
24 #9宫
25 for i in range(xi,xi+3):
26 for j in range(yj,yj+3):
27 if board[i][j]==k: return False
28 return TruesolveSudoku
=================================================================================================================
不行，脑壳疼，换哈希啃啃
242:有效的字母异位词
链接：242. 有效的字母异位词 - 力扣（LeetCode）
1 class Solution:
2 def isAnagram(self, s: str, t: str) -> bool:
3 if(len(s)!=len(t)): return False
4 for i in s:
5 if s.count(i)!=t.count(i): return False
6 return TrueisAnagram
349：两个数组的交集
链接：349. 两个数组的交集 - 力扣（LeetCode）
1 class Solution:
2 def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
3 nums1=list(set(nums1))
4 nums2=list(set(nums2))
5 re=[]
6 for i in nums1:
7 if(i in nums2): re.append(i)
8 return re intersection
202:快乐数
1 class Solution:
2 def isHappy(self, n: int) -> bool:
3 hashcode=[n]
4 sums=0
5 while True:
6 if(not n):
7 if(sums==1): return True
8 if(sums in hashcode): return False
9 hashcode.append(sums)
10 n,sums=sums,n
11 else:
12 m=n%10
13 n=n//10
14 sums+=m*misHappy