有效的数独

2024年6月10日 128次阅读

有9×9数量不变数独用二维数组表示，检查数独是否满足以下条件：

  1. 数字1-9每行只能出现一次
  2. 数字1-9每列只能出现一次
  3. 数字1-9每3×3的框内只能出现一次

数独空白部分用‘.’填充，列子如下

var a = [
  ["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"]
]

只验证上面数独是否满足要求，并不需要验证数独是否可解，代码如下所示

/**
 * @param {String[][]} board
 * @return {boolean}
 */
let isValidSudoku = function(board) {
    for (let i = 0; i < 9; i++) {
        let row = new Set();
        let column = new Set();
        let block = new Set();
        for(let j = 0;j < 9; j++) {
            if (board[i][j] !== '.') {
                let current_row = board[i][j];

                if (row.has(current_row)) {
                    console.log('row',i,j,current_row, row);
                    return false;
                }else {
                    row.add(current_row);
                }
            }

            if (board[j][i] !== '.') {
                let current_col = board[j][i];
                if (column.has(current_col)) {
                    console.log('column',i,j,current_col, column);
                    return false;
                }else {
                    column.add(current_col);
                }
            }

            let x = (+(i/3).toString()[0])*3 + (+(j/3).toString()[0]);
            let y = (i%3)*3 + j%3;
            let current_blo = board[x][y];
            if (current_blo !== '.') {
                if (block.has(current_blo)) {
                    console.log('block',i,j,current_blo, block);
                    return false;
                } else {
                    block.add(current_blo);
                }
            }
        }
    }
    return true;
};

基本思路是利用set数据结构验证数字每一行每一列以及每一块儿是否存在重复数字，其中验证块较为复杂点，需要找到对应i，j的关系，并且js中除法运算不是取整，因此要手动取整

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