蓝桥杯 第十一天 回溯法（2）
作者：互联网
目录
1.93. 复原 IP 地址 - 力扣（LeetCode） (leetcode-cn.com)
2.78. 子集 - 力扣（LeetCode） (leetcode-cn.com)
3.90. 子集 II - 力扣（LeetCode） (leetcode-cn.com)
4.491. 递增子序列 - 力扣（LeetCode） (leetcode-cn.com)
5.46. 全排列 - 力扣（LeetCode） (leetcode-cn.com)
6.47. 全排列 II - 力扣（LeetCode） (leetcode-cn.com)
7.51. N 皇后 - 力扣（LeetCode） (leetcode-cn.com)
8.37. 解数独 - 力扣（LeetCode） (leetcode-cn.com)
1.93. 复原 IP 地址 - 力扣（LeetCode） (leetcode-cn.com)
切割问题
class Solution:
def restoreIpAddresses(self, s: str) -> List[str]:
results=[]
def dfs(rest,times,path):
if times==4:
if not rest:
results.append(path[:-1])
return
for i in range(1,len(rest)+1):
if rest[:i]=='0' or (rest[0]!='0' and 0<int(rest[:i])<=255):
dfs(rest[i:],times+1,path+rest[:i]+'.')
return
dfs(s,0,"")
return resultsclass Solution:
def restoreIpAddresses(self, s: str) -> List[str]:
results=[]
def dfs(s,index,path,times):
if times==4:
if index==len(s):
results.append(path[:-1])
return
for i in range(index,len(s)):
if s[index:i+1]=='0' or (s[index]!='0' and 0<int(s[index:i+1])<=255):
dfs(s,i+1,path+s[index:i+1]+'.',times+1)
return
dfs(s,0,"",0)
return results2.78. 子集 - 力扣（LeetCode） (leetcode-cn.com)
class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
results=[]
def dfs(rest,path):
if path not in results:
results.append(path)
if not rest:
return
for i in range(len(rest)):
dfs(rest[i+1:],path+rest[:i+1])
dfs(rest[i+1:],path)
dfs(nums,[])
return resultsclass Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
results=[]
def dfs(nums,index,path):
results.append(path)
for i in range(index,len(nums)):
dfs(nums,i+1,path+[nums[i]])
dfs(nums,0,[])
return results3.90. 子集 II - 力扣（LeetCode） (leetcode-cn.com)
for循环后面的第一句话，堪称防重复的精髓
class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
results=[]
nums.sort()
def dfs(nums,index,path):
results.append(path)
for i in range(index,len(nums)):
if i>index and nums[i]==nums[i-1]:
continue
dfs(nums,i+1,path+[nums[i]])
dfs(nums,0,[])
return results4.491. 递增子序列 - 力扣（LeetCode） (leetcode-cn.com)
用哈希表防止重复
class Solution:
def findSubsequences(self, nums: List[int]) -> List[List[int]]:
results=[]
def dfs(nums,index,path):
if len(path)>=2:
results.append(path)
if index==len(nums):
return
hashtable=[0 for i in range(-101,101)]
for i in range(index,len(nums)):
# if i>index and nums[i]==nums[i-1]:
# continue
if hashtable[nums[i]]!=0:
continue
if not path:
hashtable[nums[i]]+=1
dfs(nums,i+1,path+[nums[i]])
else:
if nums[i]>=path[-1]:
hashtable[nums[i]]+=1
dfs(nums,i+1,path+[nums[i]])
return
dfs(nums,0,[])
return results5.46. 全排列 - 力扣（LeetCode） (leetcode-cn.com)
class Solution:
def permute(self, nums: List[int]) -> List[List[int]]:
results=[]
hashtable=[False for i in range(len(nums))]
def dfs(nums,path):
if len(path)==len(nums):
results.append(path)
return
for i in range(len(nums)):
if not hashtable[i]:
hashtable[i]=True
dfs(nums,path+[nums[i]])
hashtable[i]=False
return
dfs(nums,[])
return results6.47. 全排列 II - 力扣（LeetCode） (leetcode-cn.com)
一个很奇妙的bug，出现在localhash那里
class Solution:
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
results=[]
hashtable=[False for i in range(len(nums))]
def dfs(nums,path):
if len(path)==len(nums):
results.append(path)
return
localhash=[False for i in range(-11,11)]
for i in range(len(nums)):
if localhash[nums[i]]==True:
continue
if not hashtable[i]:
localhash[nums[i]]=True
hashtable[i]=True
dfs(nums,path+[nums[i]])
hashtable[i]=False
return
dfs(nums,[])
return results7.51. N 皇后 - 力扣（LeetCode） (leetcode-cn.com)
class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
result=[]
table=[-1 for i in range(n)]
def check(row,col):
if col in table:
return False
for i in range(row):
if abs(row-i)==abs(col-table[i]):
return False
return True
def dfs(level,path):
nonlocal result
if level==n:
result.append(path)
return
for i in range(n):
if check(level,i):
currow="."*i+"Q"+"."*(n-i-1)
table[level]=i
dfs(level+1,path+[currow])
table[level]=-1
dfs(0,[])
return result8.37. 解数独 - 力扣（LeetCode） (leetcode-cn.com)
大型N皇后
class Solution:
def solveSudoku(self, board: List[List[str]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
def backtrack(board):
for i in range(len(board)): #遍历行
for j in range(len(board[0])): #遍历列
if board[i][j] != ".": continue
for k in range(1,10): #(i, j) 这个位置放k是否合适
if isValid(i,j,k,board):
board[i][j] = str(k) #放置k
if backtrack(board): return True #如果找到合适一组立刻返回
board[i][j] = "." #回溯，撤销k
return False #9个数都试完了，都不行，那么就返回false
return True #遍历完没有返回false，说明找到了合适棋盘位置了
def isValid(row,col,val,board):
for i in range(9): #判断行里是否重复
if board[row][i] == str(val):
return False
for j in range(9): #判断列里是否重复
if board[j][col] == str(val):
return False
startRow = (row // 3) * 3
startcol = (col // 3) * 3
for i in range(startRow,startRow + 3): #判断9方格里是否重复
for j in range(startcol,startcol + 3):
if board[i][j] == str(val):
return False
return True
backtrack(board)标签：第十一天,return,nums,List,results,dfs,蓝桥,回溯,path 来源： https://blog.csdn.net/aixiaoxiao13/article/details/122790778