重新安排行程

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Companies

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：

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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

Discussion | Solution

题解

// @lc code=start
class Solution {
private:unordered_map<string,map<string,int>> targets;bool backtracking(int ticketNum, vector<string>& result) {if(result.size() == ticketNum + 1) {return true;}for(pair<const string,int>& target : targets[result[result.size()-1]]) {if(target.second > 0) {result.push_back(target.first);target.second--;if(backtracking(ticketNum,result)) return true;result.pop_back();target.second++;}}return false;}public:vector<string> findItinerary(vector<vector<string>>& tickets) {targets.clear();vector<string> result;for(const vector<string>& vec : tickets) {targets[vec[0]][vec[1]]++;}result.push_back("JFK");backtracking(tickets.size(),result);return result;}
};

[参考文章](代码随想录 (programmercarl.com))

N 皇后

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Companies

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

Discussion | Solution

题解

// @lc code=start
class Solution {
private:vector<vector<string>> result;void backtracking(int n,int row,vector<string>& chessboard) {if(row == n) {result.push_back(chessboard);return;}for(int col = 0; col < n; col++) {if(isValid(row,col,chessboard,n)) {chessboard[row][col] = 'Q';backtracking(n,row + 1,chessboard);chessboard[row][col] = '.'; }}}bool isValid(int row,int col,vector<string>& chessboard, int n) {for(int i = 0; i < row; ++i) {if(chessboard[i][col] == 'Q') {return false;}}for(int i = row - 1, j = col - 1; i >= 0 && j >= 0 ; i--,j--) {if(chessboard[i][j] == 'Q') {return false;}}for(int i = row - 1, j = col + 1; i >= 0 && j >=0; i--, j++) {if(chessboard[i][j] == 'Q') {return false;}}return true;}
public:vector<vector<string>> solveNQueens(int n) {result.clear();vector<string> chessboard(n,string(n,'.'));backtracking(n,0,chessboard);return result;}
};

[参考文章](代码随想录 (programmercarl.com))

解数独

Tags

数组 | 回溯 | 矩阵

Companies

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：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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

Discussion | Solution

题解

// @lc code=start
class Solution {
private:bool backtracking(vector<vector<char>>& board) {for(int i = 0; i < board.size(); i++) {for(int j = 0; j < board[0].size();j++) {if(board[i][j] == '.') {for(char k = '1'; k <= '9';k++) {if(isValid(i,j,k,board)) {board[i][j] = k;if(backtracking(board)) return true;board[i][j] = '.';}}return false;}}}return true;} bool isValid(int row,int col,char val,vector<vector<char>>& board) {for(int i = 0; i < 9;i++) {if(board[row][i] == val) {return false;}}for(int j = 0; j < 9; j++) {if(board[j][col] == val) {return false;}}int startRow = (row/3)*3;int startCol = (col/3)*3;for(int i = startRow; i < startRow + 3;i++) {for(int j = startCol; j < startCol + 3; j++) {if(board[i][j] == val) {return false;}}}return true;}
public:void solveSudoku(vector<vector<char>>& board) {backtracking(board);}
};

startCol; j < startCol + 3; j++) {
if(board[i][j] == val) {
return false;
}
}
}
return true;
}
public:
void solveSudoku(vector<vector>& board) {
backtracking(board);
}
};

[参考文章]([代码随想录 (programmercarl.com)](https://programmercarl.com/0037.解数独.html#总结))