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【LeetCode 刷题】回溯算法(5)-棋盘问题

article 2025/2/5 3:13:55

此博客为《代码随想录》二叉树章节的学习笔记，主要内容为回溯算法棋盘问题相关的题目解析。

文章目录

51. N皇后
37. 解数独
332.重新安排行程

51. N皇后

题目链接

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        board = [['.' for _ in range(n)] for _ in range(n)]
        res = []

        def check(x: int, y: int) -> bool:
            for i in range(x):
                if board[i][y] == 'Q':  # 列
                    return False
                if y - (x - i) >= 0 and board[i][y-(x-i)] == 'Q':  # 45对角线
                    return False
                if y + (x - i) < n and board[i][y+(x-i)] == 'Q':  # -45对角线
                    return False
            return True
        
        def dfs(row: int) -> None:
            if row == n:
                res.append([''.join(row) for row in board])
                return
            for col in range(n):
                if check(row, col):
                    board[row][col] = 'Q'
                    dfs(row + 1)
                    board[row][col] = '.'
        
        dfs(0)
        return res

dfs 函数的参数为行号，即“第 row 行选哪一列放皇后”构成递归树的一层
判断合法性时，不需要判断行是否合法，因为上述解法递归的逻辑即为每一行选择一列去放皇后，所以行约束肯定满足；且判断时只需要判断当前行 x 以上的合法性，因为下面的行还没有放皇后，肯定不会冲突

37. 解数独

题目链接

class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        def check(x: int, y: int, num: str) -> bool:
            for i in range(9):
                if board[i][y] == num:
                    return False
            for i in range(9):
                if board[x][i] == num:
                    return False
            row_s, col_s = x // 3 * 3, y // 3 * 3
            for i in range(3):
                for j in range(3):
                    if board[row_s+i][col_s+j] == num:
                        return False
            return True

        def dfs() -> bool:
            for i in range(9):
                for j in range(9):
                    if board[i][j] == '.':
                        for num in range(1, 10):
                            if check(i, j, str(num)):
                                board[i][j] = str(num)
                                if dfs(): return True
                                board[i][j] = '.'
                        return False
            return True
        
        dfs()
        return

数独的 check 函数需要检查整个棋盘，而非当前行列之前的内容
当找到一个解时，直接返回 True，之前的递归也不再进行回溯；如果当前位置 1-9 全部遍历也没找到解，则返回 False

332.重新安排行程

题目链接

class Solution:
    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
        tickets = sorted(tickets, key=lambda x: (x[0], x[1]))
        graph = {}
        for u, v in tickets:
            if u in graph:
                graph[u].append(v)
            else:
                graph[u] = [v]
        path = ['JFK']
        
        def dfs(u, num):
            if num == 0:
                return True
            if u not in graph or not graph[u]:
                return False
            aims = graph[u].copy()
            for v in aims:
                graph[u].pop(graph[u].index(v))
                path.append(v)
                if dfs(v, num - 1):
                    return True
                path.pop()
                graph[u].append(v)
            return False
    
        dfs("JFK", len(tickets))
        return path