LeetCode——第36题：有效的数独

时间：27ms

题目：

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

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。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 ‘.’ 表示。

示例 1:

输入:
[
[“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: true

示例 2:

输入:
[
[“8”,”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: false

解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

说明:

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
给定数独序列只包含数字 1-9 和字符 ‘.’ 。
给定数独永远是 9x9 形式的。

代码：

package leetCode;

/**
 * 2018.7.20 有效的数独
 * 笨办法，就一次判断每一列每一行和每个3x3的块
 * @author dhc
 *
 */
public class ThirtySix {
    //判断一个长度为9的一位数组是否存在相等的元素
    public static boolean judgeOneColOrRaw(char[] col) {
        boolean flag = false;
        out:for (int i = 0; i < col.length; i++) {
            if(col[i] == '.') {
                continue;
            }
            for (int j = i+1; j < col.length; j++) {
                if(col[j] == '.') {
                    continue;
                }
                if(col[i] == col[j]) {
                    flag = true;
                    break out;
                }
            }
        }
        return flag;
    }
    //获取二维数组的一列
    public static char[]  getOneCol(char[][] b,int col) {
        char[] re = new char[9];
        for (int i = 0; i < b.length; i++) {
            re[i] = b[i][col];
        }
        return re;
    }
    //获取二维数组的一行
    public static char[] getOneRaw(char[][] b,int raw) {
        char[] re = new char[9];
        for(int i = 0;i < b.length;i++) {
            re[i] = b[raw][i];
        }
        return re;
    }

    public static boolean isValidSudoku(char[][] board) {
        boolean isValid = true;
        //判断行或列
        for (int i = 0; i < 9; i++) {
            //取出第i行
            char[] raw = getOneRaw(board,i);
            //取出第i列
            char[] col = getOneCol(board,i);
            if(judgeOneColOrRaw(raw) || judgeOneColOrRaw(col)) {
                isValid = false;
                break;
            }
        }
        //判断3x3的块
        for(int i = 0;i < board.length;i = i + 3) {
            for(int j = 0;j < board.length;j = j + 3) {
                char[] tat = new char[9];
                tat[0] = board[i][j];
                tat[1] = board[i][j+1];
                tat[2] = board[i][j+2];
                tat[3] = board[i+1][j];
                tat[4] = board[i+1][j+1];
                tat[5] = board[i+1][j+2];
                tat[6] = board[i+2][j];
                tat[7] = board[i+2][j+1];
                tat[8] = board[i+2][j+2];
                if(judgeOneColOrRaw(tat)) {
                    isValid = false;
                }
            }
        }
        return isValid;
    }
    public static void main(String[] args) {
        char[][] b = new char[][]{
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},
            {'8','3','.','.','7','.','.','.','.'},};
        System.out.println(isValidSudoku(b));
    }
}