【代码随想录26】332.重新安排行程 51.N皇后 37.解数独

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332.重新安排行程

题目描述

给你一份航线列表 tickets ，其中 tickets[i] = [fromi, toi] 表示飞机出发和降落的机场地点。请你对该行程进行重新规划排序。

所有这些机票都属于一个从 JFK（肯尼迪国际机场）出发的先生，所以该行程必须从 JFK 开始。如果存在多种有效的行程，请你按字典排序返回最小的行程组合。

例如，行程 ["JFK", "LGA"] 与 ["JFK", "LGB"] 相比就更小，排序更靠前。

假定所有机票至少存在一种合理的行程。且所有的机票必须都用一次且只能用一次。

示例 1：

输入：tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
输出：["JFK","MUC","LHR","SFO","SJC"]

示例 2：

输入：tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
输出：["JFK","ATL","JFK","SFO","ATL","SFO"]
解释：另一种有效的行程是 ["JFK","SFO","ATL","JFK","ATL","SFO"] ，但是它字典排序更大更靠后。

提示：

1 <= tickets.length <= 300
tickets[i].length == 2
fromi.length == 3
toi.length == 3
fromi 和 toi 由大写英文字母组成
fromi != toi

参考代码

class Solution {Map<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>();List<String> itinerary = new LinkedList<String>();public List<String> findItinerary(List<List<String>> tickets) {for (List<String> ticket : tickets) {String src = ticket.get(0), dst = ticket.get(1);if (!map.containsKey(src)) {map.put(src, new PriorityQueue<String>());}map.get(src).offer(dst);}dfs("JFK");Collections.reverse(itinerary);return itinerary;}public void dfs(String curr) {while (map.containsKey(curr) && map.get(curr).size() > 0) {String tmp = map.get(curr).poll();dfs(tmp);}itinerary.add(curr);}
}

51.N皇后

题目描述

按照国际象棋的规则，皇后可以攻击与之处在同一行或同一列或同一斜线上的棋子。

n 皇后问题 研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。

给你一个整数 n ，返回所有不同的 n 皇后问题 的解决方案。

每一种解法包含一个不同的 n 皇后问题 的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。

示例 1：

输入：n = 4
输出：[[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
解释：如上图所示，4 皇后问题存在两个不同的解法。

示例 2：

输入：n = 1
输出：[["Q"]]

提示：

1 <= n <= 9

参考代码

class Solution {public List<List<String>> solveNQueens(int n) {List<List<String>> solutions = new ArrayList<List<String>>();int[] queens = new int[n];Arrays.fill(queens, -1);Set<Integer> columns = new HashSet<Integer>();Set<Integer> diagonals1 = new HashSet<Integer>();Set<Integer> diagonals2 = new HashSet<Integer>();backtrack(solutions, queens, n, 0, columns, diagonals1, diagonals2);return solutions;}public void backtrack(List<List<String>> solutions, int[] queens, int n, int row, Set<Integer> columns, Set<Integer> diagonals1, Set<Integer> diagonals2) {if (row == n) {List<String> board = generateBoard(queens, n);solutions.add(board);} else {for (int i = 0; i < n; i++) {if (columns.contains(i)) {continue;}int diagonal1 = row - i;if (diagonals1.contains(diagonal1)) {continue;}int diagonal2 = row + i;if (diagonals2.contains(diagonal2)) {continue;}queens[row] = i;columns.add(i);diagonals1.add(diagonal1);diagonals2.add(diagonal2);backtrack(solutions, queens, n, row + 1, columns, diagonals1, diagonals2);queens[row] = -1;columns.remove(i);diagonals1.remove(diagonal1);diagonals2.remove(diagonal2);}}}public List<String> generateBoard(int[] queens, int n) {List<String> board = new ArrayList<String>();for (int i = 0; i < n; i++) {char[] row = new char[n];Arrays.fill(row, '.');row[queens[i]] = 'Q';board.add(new String(row));}return board;}
}

37.解数独

题目描述

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1：

输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：

提示：

board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据保证输入数独仅有一个解

参考代码

class Solution {private boolean[][] line = new boolean[9][9];private boolean[][] column = new boolean[9][9];private boolean[][][] block = new boolean[3][3][9];private boolean valid = false;private List<int[]> spaces = new ArrayList<int[]>();public void solveSudoku(char[][] board) {for (int i = 0; i < 9; ++i) {for (int j = 0; j < 9; ++j) {if (board[i][j] == '.') {spaces.add(new int[]{i, j});} else {int digit = board[i][j] - '0' - 1;line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;}}}dfs(board, 0);}public void dfs(char[][] board, int pos) {if (pos == spaces.size()) {valid = true;return;}int[] space = spaces.get(pos);int i = space[0], j = space[1];for (int digit = 0; digit < 9 && !valid; ++digit) {if (!line[i][digit] && !column[j][digit] && !block[i / 3][j / 3][digit]) {line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;board[i][j] = (char) (digit + '0' + 1);dfs(board, pos + 1);line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false;}}}
}