• LeetCode No36. 有效的数独


    题目

    请你判断一个 9 x 9 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。(请参考示例图)

    注意:

    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。
    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:true

    示例 2:

    输入:board =
    [["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。

    提示:

    board.length == 9
    board[i].length == 9
    board[i][j] 是一位数字(1-9)或者 '.'

    思路

    暴力模拟

    数独的定义是行和列都是没有重复的数,那么可以在初始的时候就直接检查行和列是否满足条件,然后再每个小矩阵的起点(小矩阵左上角),查看是否有重复的数即可。

    哈希

    暴力之后看了题解才想到的原来可以用哈希的思想,将每行,每列,每个小矩阵都哈希到一个数组中,如果出现有重复的数,则是不合格的。

    AC代码

    暴力模拟

    点击查看代码
    class Solution {
        char[][] board;
    
        private boolean initCheck() {
            for(int i=0; i<9; i++) {
                boolean[] rowOk = new boolean[10];
                boolean[] colOk = new boolean[10];
                for(int j=0; j<9; j++) {
                    if(board[i][j]!='.') {
                        int numR = (int) (board[i][j] - '0');
                        if( rowOk[numR] ) {
                            return false;
                        }
                        rowOk[numR] = true;
                    }
                    if( board[j][i]!='.') {
                        int numC = (int) (board[j][i] - '0');
                        if( colOk[numC] ) {
                            return false;
                        }
                        colOk[numC] = true;
                    }
                }
            }
            return true;
        }
    
        private boolean isLittleSudo(int row, int col) {
            // System.out.println("row = " + row + " col" + col);
            boolean[] arr = new boolean[10];
            int endR = row + 3;
            int endC = col + 3;
            // System.out.println("endR = " + endR + " endC" + endC);
            for(int i=row; i<endR; i++) {
                for(int j=col; j<endC; j++) {
                    // if( row==0 && col==3 ) {
                        // System.out.print(board[i][j]);
                    // } 
                    if(board[i][j]=='.') {
                        continue;
                    }
                    int num = (int) (board[i][j] - '0');
                    if( arr[num] ) {
                        return false;
                    }
                    arr[num] = true;
                }
                // System.out.println();
            }
            return true;
        }
    
        public boolean isValidSudoku(char[][] board) {
            this.board = board;
            boolean flag = initCheck();
            if( !flag ) {
                return false;
            }
            for(int i=0; i<9; i+=3) {
                for(int j=0; j<9; j+=3) {
                    if( !isLittleSudo(i,j) ) {
                        return false;
                    }
                }
            }
            return true;
        }
    }
    

    哈希

    点击查看代码
    class Solution {
        public boolean isValidSudoku(char[][] board) {
            boolean[][] rowOk = new boolean[10][10];
            boolean[][] colOk = new boolean[10][10];
            boolean[][][] littleSudo = new boolean[3][3][10];
            for(int i=0; i<9; i++) {
                for(int j=0; j<9; j++) {
                    char ch = board[i][j];
                    if( ch == '.' ) {
                        continue;
                    }
                    int num = (int)( board[i][j]-'0');
                    if( rowOk[i][num] || colOk[j][num] || littleSudo[i/3][j/3][num] ) {
                        return false;
                    }
                    rowOk[i][num] = true;
                    colOk[j][num] = true;
                    littleSudo[i/3][j/3][num] = true;
                }
            }
            return true;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/Asimple/p/16255659.html
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