332. Reconstruct Itinerary 51. N-Queens 37. Sudoku Solver

332. Reconstruct Itinerary

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary行程 in order and return it.

All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.

• For example, the itinerary ["JFK", "LGA"] has a smaller lexical词序(B>A) order than ["JFK", "LGB"].

You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

 

infinite loop： JFK->NRT->JFK->.........

 

 

Solving graphy:

 

 using used:

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        tickets.sort() # 先排序，这样一旦找到第一个可行路径，一定是字母排序最小的
        used = [0] * len(tickets)
        path = ['JFK']  # 不是[]empty set
        results = []
        self.backtracking(tickets, used, path, 'JFK', results)
        return results[0]
    
    def backtracking(self, tickets, used, path, cur, results):
        if len(path) == len(tickets) + 1:  # 终止条件：路径长度等于机票数量+1
            results.append(path[:])  # 将当前路径添加到结果列表
            return True
        
        for i, ticket in enumerate(tickets):  # 遍历机票列表
            if ticket[0] == cur and used[i] == 0:  # 找到起始机场为cur且未使用过的机票
                used[i] = 1  # 标记该机票为已使用
                path.append(ticket[1])  # 将到达机场添加到路径中
                state = self.backtracking(tickets, used, path, ticket[1], results)  # 递归搜索
                path.pop()  # 回溯，移除最后添加的到达机场
                used[i] = 0  # 标记该机票为未使用
                if state:
                    return True  # 只要找到一个可行路径就返回，不继续搜索

using Dictionary:

from collections import defaultdict

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        targets = defaultdict(list)  # 构建机场字典
        for ticket in tickets:
            targets[ticket[0]].append(ticket[1])
        for airport in targets:
            targets[airport].sort()  # 对目的地列表进行排序

        path = ["JFK"]  # 起始机场为"JFK"
        self.backtracking(targets, path, len(tickets))
        return path

    def backtracking(self, targets, path, ticketNum):
        if len(path) == ticketNum + 1:
            return True  # 找到有效行程

        airport = path[-1]  # 当前机场
        destinations = targets[airport]  # 当前机场可以到达的目的地列表
        for i, dest in enumerate(destinations):
            targets[airport].pop(i)  # 标记已使用的机票
            path.append(dest)  # 添加目的地到路径
            if self.backtracking(targets, path, ticketNum):
                return True  # 找到有效行程
            targets[airport].insert(i, dest)  # 回溯，恢复机票
            path.pop()  # 移除目的地
        return False  # 没有找到有效行程

using Dictionaries with reverse Order:

from collections import defaultdict

class Solution:
    def findItinerary(self, tickets):
        targets = defaultdict(list)  # 创建默认字典，用于存储机场映射关系
        for ticket in tickets:
            targets[ticket[0]].append(ticket[1])  # 将机票输入到字典中
        
        for key in targets:
            targets[key].sort(reverse=True)  # 对到达机场列表进行字母逆序排序
        
        result = []
        self.backtracking("JFK", targets, result)  # 调用回溯函数开始搜索路径
        return result[::-1]  # 返回逆序的行程路径
    
    def backtracking(self, airport, targets, result):
        while targets[airport]:  # 当机场还有可到达的机场时
            next_airport = targets[airport].pop()  # 弹出下一个机场
            self.backtracking(next_airport, targets, result)  # 递归调用回溯函数进行深度优先搜索
        result.append(airport)  # 将当前机场添加到行程路径中

51. N-Queens

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

Given an integer n, return all distinct不同的 solutions to the n-queens puzzle. You may return the answer in any order.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space, respectively.

 Queen's constraints：

1.Can't be on the same row

2.Can't be in the same column

3.Can't be the same diagonal 对角线 /daɪˈæɡənl/

 

 了解backtracking，其实要用DFS

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        result = []  # 存储最终结果的二维字符串数组

        chessboard = ['.' * n for _ in range(n)]  # 初始化棋盘
        self.backtracking(n, 0, chessboard, result)  # 回溯求解
        return [[''.join(row) for row in solution] for solution in result]  # 返回结果集

    def backtracking(self, n: int, row: int, chessboard: List[str], result: List[List[str]]) -> None:
        if row == n:
            result.append(chessboard[:])  # 棋盘填满，将当前解加入结果集
            return

        for col in range(n):
            if self.isValid(row, col, chessboard):
                chessboard[row] = chessboard[row][:col] + 'Q' + chessboard[row][col+1:]  # 放置皇后
                self.backtracking(n, row + 1, chessboard, result)  # 递归到下一行
                chessboard[row] = chessboard[row][:col] + '.' + chessboard[row][col+1:]  # 回溯，撤销当前位置的皇后

    def isValid(self, row: int, col: int, chessboard: List[str]) -> bool:
        # 检查列
        for i in range(row):
            if chessboard[i][col] == 'Q':
                return False  # 当前列已经存在皇后，不合法

        # 检查 45 度角是否有皇后
        i, j = row - 1, col - 1
        while i >= 0 and j >= 0:
            if chessboard[i][j] == 'Q':
                return False  # 左上方向已经存在皇后，不合法
            i -= 1
            j -= 1

        # 检查 135 度角是否有皇后
        i, j = row - 1, col + 1
        while i >= 0 and j < len(chessboard):
            if chessboard[i][j] == 'Q':
                return False  # 右上方向已经存在皇后，不合法
            i -= 1
            j += 1

        return True  # 当前位置合法

 37. Sudoku Solver

 

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

1. Each of the digits 1-9 must occur exactly once in each row.
2. Each of the digits 1-9 must occur exactly once in each column.
3. Each of the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

The '.' character indicates empty cells.

 

 analyse graph：

There are three constraints to determine whether a board is right or not as follows:

• Whether in the same row has duplicate
• Whether in the same column has duplicate
• Whether in the 9-gallery has duplicate

backtracking:

class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        self.backtracking(board)

    def backtracking(self, board: List[List[str]]) -> bool:
        # 若有解，返回True；若无解，返回False
        for i in range(len(board)): # 遍历行
            for j in range(len(board[0])):  # 遍历列
                # 若空格内已有数字，跳过
                if board[i][j] != '.': continue
                for k in range(1, 10):
                    if self.is_valid(i, j, k, board):
                        board[i][j] = str(k)
                        if self.backtracking(board): return True
                        board[i][j] = '.'
                # 若数字1-9都不能成功填入空格，返回False无解
                return False
        return True # 有解

    def is_valid(self, row: int, col: int, val: int, board: List[List[str]]) -> bool:
        # 判断同一行是否冲突
        for i in range(9):
            if board[row][i] == str(val):
                return False
        # 判断同一列是否冲突
        for j in range(9):
            if board[j][col] == str(val):
                return False
        # 判断同一九宫格是否有冲突
        start_row = (row // 3) * 3
        start_col = (col // 3) * 3
        for i in range(start_row, start_row + 3):
            for j in range(start_col, start_col + 3):
                if board[i][j] == str(val):
                    return False
        return True

DFS:

class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        def dfs(pos: int):
            nonlocal valid
            if pos == len(spaces):
                valid = True
                return
            
            i, j = spaces[pos]
            for digit in range(9):
                if line[i][digit] == column[j][digit] == block[i // 3][j // 3][digit] == False:
                    line[i][digit] = column[j][digit] = block[i // 3][j // 3][digit] = True
                    board[i][j] = str(digit + 1)
                    dfs(pos + 1)
                    line[i][digit] = column[j][digit] = block[i // 3][j // 3][digit] = False
                if valid:
                    return
            
        line = [[False] * 9 for _ in range(9)]
        column = [[False] * 9 for _ in range(9)]
        block = [[[False] * 9 for _a in range(3)] for _b in range(3)]
        valid = False
        spaces = list()

        for i in range(9):
            for j in range(9):
                if board[i][j] == ".":
                    spaces.append((i, j))
                else:
                    digit = int(board[i][j]) - 1
                    line[i][digit] = column[j][digit] = block[i // 3][j // 3][digit] = True

        dfs(0)