Leetcode刷题第七天-回溯-哈希

332:重新岸炮行程

链接：332. 重新安排行程 - 力扣（LeetCode）

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 False

findItinerary

51：N皇后

链接：51. N 皇后 - 力扣（LeetCode）

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 re

solveNQueens

37:解数独

链接：37. 解数独 - 力扣（LeetCode）

(⊙﹏⊙)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 True

solveSudoku

=================================================================================================================

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 True

isAnagram

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

链接：202. 快乐数 - 力扣（LeetCode）

 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*m

isHappy