leetcode37. 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"]]

使用回溯大法:

public class Demo1 {
    public static void main(String[] args) {
        char[][] chars = {{'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'}};
        //char[][] chars ={{'1','.','3'}, {'.','3','.'}, {'3','.','.'}};
        new Solution().solveSudoku(chars);
    }
}


class Solution {
    public void solveSudoku(char[][] board) {
        dfs(board, 0, 0);
    }

    private boolean dfs(char[][] board, int i, int j) {
        if (i == board.length  || j == board[i].length ) {
            return true;
        }
        char cur = board[i][j];
        if (cur == '.') {
            boolean flag = false;
            for (int k = 1; k <= 9; k++) {
                //合法
                if (isValid(i, j, k, board)) {
                    board[i][j] = (char) (k + 48);
                    //print(board);
                    if (j < board[i].length - 1) {
                        flag = dfs(board, i, j+1);
                    } else {
                        flag =dfs(board, i + 1, 0);
                    }
                    if (flag) {
                        return true;
                    } else {
                        //回溯
                        board[i][j] = '.';
                    }

                }
            }

        } else {
            if (j < board[i].length - 1) {
                return dfs(board, i, j+1);
            } else {
                return dfs(board, i + 1, 0);
            }
        }
        return false;
    }

    /**
     * 判断位置是否合法
     * @param i
     * @param j
     * @param k
     * @param arr
     * @return
     */
    private boolean isValid(int i, int j, int k, char[][] arr) {
        //行比较
        char[] rowArr = arr[i];
        for (int x = 0; x < arr[0].length; x++) {
            if (x != j && rowArr[x] == (char) (k + 48)) {
                return false;
            }
        }
        //列比较
        for (int y = 0; y < arr.length; y++) {
            if (y != i && arr[y][j] == (char) (k + 48)) {
                return false;
            }
        }
        //九宫比较

        int s1 = (i / 3) * 3;
        int s2 = (j / 3) * 3;

        for (int p = s1; p < s1 + 3; p++) {
            for (int q = s2; q < s2 + 3; q++) {
                if (arr[p][q] == (char) (k + 48)) {
                    return false;
                }
            }
        }
        return true;
    }
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值