leetcode 有效的数独

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

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

巴特西

leetcode 有效的数独

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