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

 状态：完成

思路：还是用回溯去解决这个问题，一行就是一个树层，每个树层先判断在这个位置上符不符合要求然后再进入下一层，就这样子到最后一层为止，把结果装进result数组里面。

class Solution {
    List<List<String>> result=new ArrayList<>();
    List<String> list=new LinkedList<>();
    int Q_num=0;
    public List<List<String>> solveNQueens(int n) {
        char[] arr=new char[n];
        Arrays.fill(arr,'.');
        for(int i=0;i<n;i++){
            list.add(String.valueOf(arr));
        }
        backstracking(0,0,n);
        return result;
    }
    public void backstracking(int x,int y,int n){
        if(Q_num==n){
            result.add(new LinkedList(list));
            return;
        }
        String temp=list.get(y);
        for(int i=x;i<n;i++){
            boolean flag=isVaild(i,y,n);
            System.out.println(i+" "+y+" "+flag+" "+list);
            if(flag){
                Q_num++;
                list.set(y,temp.substring(0,i)+"Q"+temp.substring(i+1,n));
                backstracking(0,y+1,n);
                list.set(y,temp.substring(0,i)+"."+temp.substring(i+1,n));
                Q_num--;
            }
        }
    }
    public boolean isVaild(int x,int y,int n){
        //列判断
        for(int i=0;i<n;i++){
            String temp=list.get(i);
            if(temp.charAt(x)=='Q') return false;
        }
        //行判断
        String temp=list.get(y);
        for(int i=0;i<n;i++){
            if(temp.charAt(i)=='Q') return false;
        }
        //斜线判断
        int startLeft_x=x;
        int startLeft_y=y;
        while(startLeft_x>0&&startLeft_y>0){
            startLeft_x--;
            startLeft_y--;
        }
        int startRight_x=x;
        int startRight_y=y;
        while(startRight_x<n-1&&startRight_y>0){
            startRight_x++;
            startRight_y--;
        }
        while(startLeft_x>=0&&startLeft_y>=0&&startLeft_x<n&&startLeft_y<n){
            String a=list.get(startLeft_y);
            if(a.charAt(startLeft_x)=='Q') return false;
            startLeft_x++;
            startLeft_y++;
        }
        while(startRight_x>=0&&startRight_y<n){
            String a=list.get(startRight_y);
            if(a.charAt(startRight_x)=='Q') return false;
            startRight_x--;
            startRight_y++;
        }
        return true;
    }
}

 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

状态：完成

思路：这里我自己写是用了暴力的递归的方式去解决问题的，就是判断合不合适合适就填进去不合适就继续下个数，一个格子一个格子的去找的，carl那种二维递归就是一行一行地确定的，每次回溯都用两个for循环。

class Solution {
    List<List<Integer>> list=new ArrayList<>();
    public void solveSudoku(char[][] board) {
        for(int i=0;i<9;i++){
            for(int k=0;k<9;k++){
                if(board[i][k]=='.'){
                    ArrayList<Integer> temp= new ArrayList<>();
                    temp.add(i);
                    temp.add(k);
                    list.add(temp);
                }
            }
        }
        System.out.println(backstracking(board,0));
    }
    public boolean backstracking(char[][] board,int index){
        if(index==list.size()) return true;
        List<Integer> temp=list.get(index);
        int x=temp.get(0);
        int y=temp.get(1);
        for(int i=1;i<=9;i++){
            if(isVaild(x,y,i,board)){
                board[x][y]=(char)(i+48);
                System.out.println(x+" "+y+" "+(char)(i+48));
                if(backstracking(board,index+1)) return true;
                board[x][y]='.';
            }
        }
        return false;
    }
    public boolean isVaild(int x,int y,int value,char[][] board){
        //9宫格检索
        int start_x=(x/3)*3;
        int start_y=(y/3)*3;
        for(int i=0;i<3;i++){
            for(int k=0;k<3;k++){
                if(board[start_x+i][start_y+k]!='.')
                    if(Integer.valueOf(String.valueOf(board[start_x+i][start_y+k]))==value){
                        return false;
                    }
            }
        }
        //行
        for(int i=0;i<9;i++){
            if(board[i][y]!='.')
                if(Integer.valueOf(String.valueOf(board[i][y]))==value){
                    return false;
                }
        }
        //列
        for(int i=0;i<9;i++){
            if(board[x][i]!='.')
                if(Integer.valueOf(String.valueOf(board[x][i]))==value){
                    return false;
                }
        }
        return true;
    }
}

 感想：终于刷完回溯了，继续进步，贪心出发。