代码随想录算法训练营第30天|32. 重新安排行程，51. N 皇后，37. 解数独-优快云博客

本文链接：https://blog.youkuaiyun.com/dragonccl/article/details/132557667

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

做这道题之前需要对图论有一定了解，那个链式建立的图表，一定要很清晰，然后这段代码中使用unordered_map<string, map<string, int>> 这种数据结构来代表图的这种链表还有防止闭环导致的无限循环，很巧妙。以后可以经常使用这种方式；

另外一点需要说明的是，在map里first是按字典顺序存储的，这个结合这段代码相信就可以很好理解本题了

class Solution {
    unordered_map<string, map<string, int>> targets;
    bool backTracking(int count,vector<string>& result){
        if(result.size()==count+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(count,result))return true;
                result.pop_back();
                target.second++;
            }
        }
        return false;
    }
public:
    vector<string> findItinerary(vector<vector<string>>& tickets) {
        vector<string> result;
        for(const vector<string>&ve:tickets){
            targets[ve[0]][ve[1]]++;
        }
        result.push_back("JFK");
        backTracking(tickets.size(),result);
        return result;
    }
};

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

n皇后问题并不复杂，主要是要搞懂如何暴力枚举，然后如何去除不合适的结果

class Solution {
    vector<vector<string>> ans;
    bool isvaild(int row,int m,vector<string> &path,int n){
        for(int i=0;i<row;i++){
            if(path[i][m]=='Q')return false;
        }
        for(int i=row-1,j=m-1;i>=0&&j>=0;i--,j--){
            if(path[i][j]=='Q')return false;
        }
        for(int i=row-1,j=m+1;i>=0&&j<n;i--,j++){
            if(path[i][j]=='Q')return false;
        }
        return true;
    }
    void backtracking(int n,int row,vector<string> &path){
        if(n==row){
            ans.push_back(path);
            return;
        }
        for(int i=0;i<n;i++){
            if(isvaild(row,i,path,n)){
                path[row][i]='Q';
                backtracking(n,row+1,path);
                path[row][i]='.';
            }
        }
    }
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<string> ve;
        for(int i=0;i<n;i++){
            string m="";
            for(int j=0;j<n;j++){
                m+=".";
            }
            ve.push_back(m);
        }
        backtracking(n,0,ve);
        return ans;
    }
};

37. 解数独

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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

这道题卡哥用了一种二维回溯法，其实也就是把二维数组展开成一位数组，等到最后一个填充结束，return true，通过bool，一直回溯到开头，直到程序运行结束

class Solution {
    bool isvaild(int row,int col,char m,vector<vector<char>> board){
        for(int i=0;i<9;i++){
            if(board[row][i]==m||board[i][col]==m)return false;
        }
        int a = (row/3); int b=(col/3);
        for(int i=a*3;i<a*3+3;i++){
            for(int j=b*3;j<b*3+3;j++){
                if(board[i][j]==m)return false;
            }
        }
        return true;
    }
    bool backTracking(vector<vector<char>>& board){
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                if(board[i][j]=='.'){
                    for(char k='1';k<='9';k++){
                        if(isvaild(i,j,k,board)){
                            board[i][j]=k;
                            if(backTracking(board))return true;
                            board[i][j]='.';
                        }
                    }
                    return false;
                }
            }
        }
        return true;
    }
public:
    void solveSudoku(vector<vector<char>>& board) {
        backTracking(board);
    }
};