代码随想录算法训练营第30天| 332.重新安排行程 、51. N皇后 、 37. 解数独

332.重新安排行程

开始想的是将行程进行全排列之后，然后选出一个字典排序最小的。就也是使用的回溯的思路。

class Solution {
private:vector<vector<string>> res;vector<string> path;int ticketUsed = 0;bool cmp(const vector<string> &a, const vector<string> &b) {for (int i = 0; i < b.size(); ++i) {if (a[i] == b[i]) continue; //跳出循环return a[i] < b[i] ? true : false;}return false;}void backtracking(vector<vector<string>> &tickets, vector<bool>& used, string airport) {if (ticketUsed == tickets.size()) {res.push_back(path);return;}for (int i = 0; i < tickets.size(); i++) {if (tickets[i][0] == airport && used[i] == false) {used[i] = true;ticketUsed++;path.push_back(tickets[i][1]);backtracking(tickets, used, tickets[i][1]);used[i] = false;ticketUsed--;path.pop_back();}}}public:vector<string> findItinerary(vector<vector<string>>& tickets) {vector<bool> used(tickets.size(), false);path.push_back("JFK");backtracking(tickets, used, "JFK");sort(res.begin(), res.end(), cmp);return res[0];}
};

使用了其他的数据结构进行去重

class Solution {
private:// unordered_map<出发机场, map<到达机场, 航班次数>> targetsunordered_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> res;for(const vector<string>& vec : tickets) {targets[vec[0]][vec[1]]++;}res.push_back("JFK");  //起始机场backtracking(tickets.size(), res);return res;}
};

51. N皇后

本题是使用回溯法的经典问题。

里面有一步就是验证放了之后，是否有冲突。 因为是从上往下一层层的在填充，所以就不用考虑 更下一层是否会有皇后。而且，也不需要考虑行

bool inValid(int row, int col, vector<string>& chessboard, int n) {// 检查列for (int i = 0; i < row; ++i) {if (cheese[i][col] == 'Q')return false;}// 135° 检查是否有皇后for (int i = row - 1, j = col - 1; i >=0 && j >= 0; --i, --j) {if (cheese[i][col] == 'Q')return false;}// 45°检查是否有皇后for (int i = row - 1, j = col + 1; i >= 0 && j <= n -1; --i, ++j) {if (cheese[i][col] == 'Q')return false;}return true;
}

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)) { // 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++) {if(chessboard[i][col] == 'Q') {return false;}}//135° 检查是否有皇后for(int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {if(chessboard[i][j] == 'Q') {return false;}}//45°检查是否有皇后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();vector<string> chessboard(n, string(n, '.'));backtracking(n, 0, chessboard);return result;}
};

37. 解数独

在看了n皇后之后，再看数独，这种填充问题，感觉就是得用回溯法。

横向的for循环来控制要填什么，纵向的递归就是在寻找一个可以填的空位。

class Solution {
private:bool isValid(int row, int col, char val, vector<vector<char>>& board) {// 一行里面是否有重复for (int i = 0; i < 9; ++i) {if (board[row][9] == val)return false;}// 一列里面是否有重复for (int i = 0; i < 9; ++i) {if (board[i][col] == val)return false;}// 判断九方格里面是否有重复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;}bool backtracking(vector<vector<char>>& board) {// 要遍历完整个树，所以不设置终止条件for (int i = 0; i < board.size(); ++i) {for (int j = 0; j < board.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) == true)return true;board[i][j] = '.'; //回溯}}// 9个数字都不行return false;}}}return true;}
public:void solveSudoku(vector<vector<char>>& board) {backtracking(board);}
};