LeetCode刷题day15——回溯法(end)

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

代码：

刚开始看到这题是困难，我都不想做了。。。但是课堂上偏偏用的是N皇后讲回溯法，没事，我还能肝！下一道是解数独，也是困难，但是我对这个蛮感兴趣的，那就做一做，看看计算机是怎么思考的！肝！

这题是用横向循环遍历行内元素，递归深度遍历每一行。一层循环就行。一开始看到棋盘问题，就会觉得是二重循环，其实不是，因为每层只需要填一个数字而已。

蛮有意思的，这题！

class Solution {
public:
    vector<vector<string>> res;
    //set<vector<string>> res1;
    bool isval(const vector<string> chessboard,int r,int c,int n) {
        //这里并不用验证行，是因为每次行是递归的，在验证时，行上肯定只有一个Q
        //col
        for(int i=0;i<n;i++)
            if(chessboard[i][c]=='Q') return false;;
        //45度
        for(int i=r-1,j=c+1;i>=0&&j<n;i--,j++) {
            if(chessboard[i][j]=='Q') return false;;
        }
        //这样写有bug，因为棋盘大的时候，斜对角不止一个格子
        // if(r-1>=0&&c+1<n&&chessboard[r-1][c+1]=='Q') return false;
        // if(r+1<n&&c-1>=0&&chessboard[r+1][c-1]=='Q') return false;
        //135
        for(int i=r-1,j=c-1;i>=0&&j>=0;i--,j--) {
            if(chessboard[i][j]=='Q') return false;;
        }
        //if(r-1>=0&&c-1>=0&&chessboard[r-1][c-1]=='Q') return false;
        // if(r+1<n&&c+1<n&&chessboard[r+1][c+1]=='Q') return false;
        return true;
    }
    void backtracking(vector<string> &chessboard,int n, int r) {
        if(r==n) {//注意这里的==，因为r from 0 to n-1
            res.push_back(chessboard);
            return;
        }
        for(int col=0;col<n;col++) {
            if(isval(chessboard,r,col,n)) {
                chessboard[r][col]='Q';
                backtracking(chessboard,n,r+1);
                chessboard[r][col]='.';//back
            }
        }
    }

    vector<vector<string>> solveNQueens(int n) {
       // vector<string> chessboard(n,(n,"."));
       //一直有很多重复结果，其实去重之后答案也并不正确，棋盘初始化的错误导致，输出了很多错误的结果
        vector<string> chessboard(n, string(n, '.'));

        backtracking(chessboard,n,0);
        //以后最好从过程中找问题，而不是最后来处理重复的结果，本质不在这里
        //res.assign(res1.begin(),res1.end());
        return res;
    }
};

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：

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

代码：

这是我做到的第一个需要在递归里面用两层循环的题，而且上来就写循环，也没有终止条件？！是的，总会酷酷退出的。本题有难度，写！

class Solution {
public:
    bool isval(vector<vector<char>> & board, int r, int c, char val) {
        //row
        for (int i = 0; i < 9; i++) {
            if(board[r][i] == val)
                return false;
            if(board[i][c] ==val)
                return false;
        }
        int start_row=(r/3)*3;//只用检查自己所在的九宫格即可，这样直接算出来挺好
        int start_col=(c/3)*3;
        for (int i = start_row; i < start_row + 3; i++) {
            for (int j = start_col; j < start_col + 3; j++) {
                if(board[i][j] == val)
                    return false;
            }
        }
        return true;
        
    }

    bool backtracking(vector<vector<char>> & board) {
        //此函数没有终止条件，总会终止的
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if(board[i][j] != '.') continue;//这里忘记跳过了
                for (char k = '1'; k <='9'; k++) {//字符1-9竟然还可以这样，以前没用过这样的
                    if(isval(board,i,j,k)) {
                        board[i][j] = k;
                        if(backtracking(board)) return true;
                        board[i][j] = '.';
                    }
                }
                return false;//九个数字都不行，肯定不行
            }
        }
        return true;
    }

    void solveSudoku(vector<vector<char>>& board) {
        backtracking(board);
    }
};