【经典LeetCode算法题目专栏分类】【第3期】回溯问题系列：单词搜索、N皇后问题、判断有效数独、解数独

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单词搜索

class Solution:     def exist(self, board: List[List[str]], word: str) -> bool:         self.m = len(board)         self.n = len(board[0])         for i in range(self.m):             for j in range(self.n):                 if board[i][j] == word[0]:                     if self.check(board,i,j,word, 0):                         return True         return False          def check(self, board, i, j, word, index):         if i < 0 or i >= self.m or j < 0 or j >= self.n or board[i][j] != word[index]:             return False         if index == len(word) - 1:             return True         # 把当前坐标的值保存下来，为了在最后复原,回溯         t = board[i][j]         # 然后修改当前坐标的值，避免之前找到过的元素又被找到一次         board[i][j] = '.'         res = self.check(board,i,j-1,word,index + 1) or self.check(board,i,j+1,word,index + 1) or self.check(board,i-1,j,word,index + 1) or self.check(board,i+1,j,word,index + 1)               board[i][j] = t         return res

N皇后问题

class Solution:     def solveNQueens(self, n: int) -> List[List[str]]:         def isValid(row, col):             for i in range(row):                 for j in range(n):                     # 注：左斜对角线上，同一条斜线上的每个位置满足行下标与列下标之差相等                     # 注：右斜对角线上，同一条斜线上的每个位置满足行下标与列下标之和相等                     if board[i][j] == 'Q' and (j == col or i + j == row + col or i-j == row-col):                         return False             return True         def backtrack(board, row):             if row >= n:                 cur_res = [''.join(row) for row in board]                 res.append(cur_res)                 return             for i in range(n):                 if isValid(row, i, board):                     board[row][i] = 'Q'                     backtrack(board, row+1)                     board[row][i] = '.'         res = []         board = [['.'] * n for _ in range(n)]         backtrack(board,0)         return res # 优化，通过集合记录之前放置过元素的正向对角线，负向对角线，及列，判断当前点是否在集合中，在的话说明不满足要求 def solveNQueens(self, n: int) -> List[List[str]]:         def isValid(row, col):             # 注：左斜对角线上，同一条斜线上的每个位置满足行下标与列下标之差相等             # 注：右斜对角线上，同一条斜线上的每个位置满足行下标与列下标之和相等             if col in col_hash or (row + col) in pie_hash or (row-col) in na_hash:                 return False             return True         def backtrack(board, row):             if row >= n:                 cur_res = [''.join(row) for row in board]                 res.append(cur_res)                 return             for col in range(n):                 if isValid(row, col, board):                     board[row][col] = 'Q'                     pie_hash.add(row + col)                     na_hash.add(row - col)                     col_hash.add(col)                     backtrack(board, row+1)                     board[row][col] = '.'                     pie_hash.remove(row + col)                     na_hash.remove(row-col)                     col_hash.remove(col)         res = []         board = [['.'] * n for _ in range(n)]         pie_hash = set()         na_hash = set()         col_hash = set()         backtrack(board,0)         return res

判断有效数独

class Solution:     def isValidSudoku(self, board: List[List[str]]) -> bool:         row_set = [set() for _ in range(9)]         col_set = [set() for _ in range(9)]         square_set = [[set() for _ in range(3)] for _ in range(3)]  #3*3         for i in range(9):             for j in range(9):                 if board[i][j] in row_set[i] or board[i][j] in col_set[j] or board[i][j] in square_set[i//3][j//3]:                     return False                 if board[i][j] != '.':                     row_set[i].add(board[i][j])                     col_set[j].add(board[i][j])                     square_set[i//3][j//3].add(board[i][j])         return True

解数独

解法一

class Solution:     def solveSudoku(self, board: List[List[str]]) -> None:         """         Do not return anything, modify board in-place instead.         """         nums = {"1", "2", "3", "4", "5", "6", "7", "8", "9"}         row = [set() for _ in range(9)]         col = [set() for _ in range(9)]         palace = [[set() for _ in range(3)] for _ in range(3)]  # 3*3         blank = []         # 初始化，按照行、列、宫 分别存入哈希表         for i in range(9):             for j in range(9):                 ch = board[i][j]                 if ch == ".":                     blank.append((i, j))                 else:                     row[i].add(ch)                     col[j].add(ch)                     palace[i//3][j//3].add(ch)         def dfs(n):             if n == len(blank):                 return True             i, j = blank[n]             rst = nums - row[i] - col[j] - palace[i//3][j//3]  # 剩余的数字             ### rst = nums - (row[i] | col[j] | palace[i//3][j//3])               if not rst:                 return False             for num in rst:                 board[i][j] = num                 row[i].add(num)                 col[j].add(num)                 palace[i//3][j//3].add(num)                 if dfs(n+1):                     return True                 row[i].remove(num)                 col[j].remove(num)                 palace[i//3][j//3].remove(num)         dfs(0)

解法二

class Solution:     def solveSudoku(self, board: List[List[str]]) -> None:         """         Do not return anything, modify board in-place instead.         """         self.board = board         self.solve()          def findUnassigned(self):         for row in range(9):             for col in range(9):                 if self.board[row][col] == ".":                     return row, col         return -1, -1          def solve(self):         row, col = self.findUnassigned()         #no unassigned position is found, puzzle solved         if row == -1 and col == -1:             return True         for num in ["1","2","3","4","5","6","7","8","9"]:             if self.isSafe(row, col, num):                 self.board[row][col] = num                 if self.solve():                     return True                 self.board[row][col] = "."         return False                  def isSafe(self, row, col, ch):         boxrow = row - row%3         boxcol = col - col%3         if self.checkrow(row,ch) and self.checkcol(col,ch) and self.checksquare(boxrow, boxcol, ch):             return True         return False          def checkrow(self, row, ch):         for col in range(9):             if self.board[row][col] == ch:                 return False         return True          def checkcol(self, col, ch):         for row in range(9):             if self.board[row][col] == ch:                 return False         return True             def checksquare(self, row, col, ch):         for r in range(row, row+3):             for c in range(col, col+3):                 if self.board[r][c] == ch:                     return False         return True

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