leetcode 有效的数独

Posted 2020-12-25 sbzy

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。

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。
   

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

看见这道题我觉得应该能做出来，因为我做过五子棋，也是自己研究的数组棋盘遍历，感觉差不多
然而，没想到，做出来就用了我小半个下午的时间，并且用的也是很笨的遍历办法…

var isValidSudoku = function (board) {
    //横向遍历
    for (let i = 0; i !== 9; i++) {
        let temp = board[i].filter(value => value !== ".");
        if (new Set(temp).size !== temp.length) {
            return false;
        }
    }
    //纵向遍历
    for (let i = 0; i !== 9; i++) {
        let temp = board.map(value => value.filter((value, index) => index === i)).map(value => value[0]).filter(value => value !== ".");
        if (new Set(temp).size !== temp.length) {
            return false;
        }
    }
    //3x3宫格遍历
    for (let i = 0; i < 9; i += 3) {
        //先分为3个9x3的
        let temp = board.map(value => value.filter((value, index) => index >= i && index < i + 3)).filter(value => value !== ".");
        //再把9x3分成3个3x3
        for (let j = 0; j < 9; j += 3) {
            //转为1维数组去重判断长度
            let temp_3x3 = temp.filter((value, index) => index >= j && index < j + 3).join(",").split(",").filter(value => value !== ".");
            if (new Set(temp_3x3).size !== temp_3x3.length) {
                return false;
            }
        }
    }
    return true;
};

方法是很笨的硬莽干