题目:332. 重新安排行程
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
思考历程与知识点:
这道题目有几个难点:
一个行程中,如果航班处理不好容易变成一个圈,成为死循环
有多种解法,字母序靠前排在前面,让很多同学望而退步,如何该记录映射关系呢 ?
使用回溯法(也可以说深搜) 的话,那么终止条件是什么呢?
搜索的过程中,如何遍历一个机场所对应的所有机场。
题解:
class Solution { private: // unordered_map<出发机场, map<到达机场, 航班次数>> targets 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 ) { // 记录到达机场是否飞过了 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]]++; // 记录映射关系 } result.push_back("JFK"); // 起始机场 backtracking(tickets.size(), result); return result; } };
题目:51. N 皇后
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
题解:
class Solution { private: vector<vector<string>> result; // n 为输入的棋盘大小 // row 是当前递归到棋盘的第几行了 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)) { // 验证合法就可以放 chessboard[row][col] = 'Q'; // 放置皇后 backtracking(n, row + 1, chessboard); chessboard[row][col] = '.'; // 回溯,撤销皇后 } } } bool isValid(int row, int col, vector<string>& chessboard, int n) { // 检查列 for (int i = 0; i < row; i++) { // 这是一个剪枝 if (chessboard[i][col] == 'Q') { return false; } } // 检查 45度角是否有皇后 for (int i = row - 1, j = col - 1; i >=0 && j >= 0; i--, j--) { if (chessboard[i][j] == 'Q') { return false; } } // 检查 135度角是否有皇后 for(int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) { if (chessboard[i][j] == 'Q') { return false; } } return true; } public: vector<vector<string>> solveNQueens(int n) { result.clear(); std::vector<std::string> chessboard(n, std::string(n, '.')); backtracking(n, 0, chessboard); return result; } };
题目:37. 解数独
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
思考历程与知识点:
一个for循环遍历棋盘的行,一个for循环遍历棋盘的列,一行一列确定下来之后,递归遍历这个位置放9个数字的可能性
题解:
class Solution { private: bool backtracking(vector<vector<char>>& board) { for (int i = 0; i < board.size(); i++) { // 遍历行 for (int j = 0; j < board[0].size(); j++) { // 遍历列 if (board[i][j] == '.') { for (char k = '1'; k <= '9'; k++) { // (i, j) 这个位置放k是否合适 if (isValid(i, j, k, board)) { board[i][j] = k; // 放置k if (backtracking(board)) return true; // 如果找到合适一组立刻返回 board[i][j] = '.'; // 回溯,撤销k } } return false; // 9个数都试完了,都不行,那么就返回false } } } return true; // 遍历完没有返回false,说明找到了合适棋盘位置了 } bool isValid(int row, int col, char val, vector<vector<char>>& board) { for (int i = 0; i < 9; i++) { // 判断行里是否重复 if (board[row][i] == val) { return false; } } for (int j = 0; j < 9; j++) { // 判断列里是否重复 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++) { // 判断9方格里是否重复 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); } };