算法(TS)：有效的数独

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2024年04月07日 18:56 · 阅读数 5

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）

算法(TS)：有效的数独

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

示例 1：


[["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：


[["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。

解答一

首先定义一个判断一维数组中是否保持重复元素的函数。遍历 board 中的每行，每列即9个 3*3 格子转成一维数组，判断这些数组中是否存在重复的元素，存在则说明 board 不是有效的数独，否则则说明board 是有效的数独。

function isValidSudoku(board: string[][]): boolean {
    // 判断数组中是否包含重复的值
    const containsDuplicate = (nums: string[],exclude:string): boolean => {
        let result: boolean = false;
        const valueSet = new Set<string>()
        let i = nums.length - 1
        while(i>-1) {
            if(nums[i] !== exclude) {
                if (valueSet.has(nums[i])) {
                    console.log(valueSet,nums)
                    result = true
                    break
                }
                valueSet.add(nums[i])
            }
            i--
        }

        return result
    }

    // 判断每行是否有重复的值
    let row = 0
    while(row < board.length) {
        if (containsDuplicate(board[row],'.')) {
            return false
        }
        row ++
    }

    // 判断每列是否有重复的值
    let col = 0
    while(col < board[0].length) {
        const colArr: string[] = []
        for(let rowIndex = 0; rowIndex < board.length; rowIndex++) {
            colArr.push(board[rowIndex][col])
        }

        if (containsDuplicate(colArr,'.')) {
            return false
        }
        col++
    }

    // 判断在 3*3 格子中是否有重复的数字
    let minBoardRow = 0, minBoardCol = 0
    while(minBoardRow < board.length / 3 && minBoardCol < board[0].length / 3) {
        const minMatrixArr: string[] = []
        let row = 0,col = 0
        while(row < 3 && col < 3) {
            minMatrixArr.push(board[row + minBoardRow * 3][col + minBoardCol * 3])
            if (col === 2) {
                row++
                col = 0
            } else {
                col++
            }
        }
        if (containsDuplicate(minMatrixArr,'.')) {
            return false
        }
        // 转到下一个 3*3 格子
        if (minBoardCol === board[0].length / 3 - 1) {
            minBoardRow++
            minBoardCol = 0
        } else {
            minBoardCol++
        }
    }

    return true
};

解答二

创建一个二维数组 rows 保存第 i 行中数字 n 出现的次数，例如 rows[1][3] = 1 表示第 2 行数字 4 出现了 1 次；创建一个二维数组 columns 保存第 j 列中数字 n 出现的次数，例如 columns[1][3] = 1 表示第 2 列数字 4 出现的 1 次

9 x 9 的数独有 9 个 3 * 3 的小九宫格，创建一个三维数组 subboxes 保存第 i 行第 j 列的小九宫格中数字 n 出现的次数，例如 subboxes[1][1][3] = 1 表示第 2 列第 2 行数字 4 出现了 1 次。

9 * 9 数独的下标 i 和 j 对应小九宫格的下标为 [⌊3/i⌋][⌊3/j⌋]。

function isValidSudoku(board: string[][]): boolean {
    const rows: number[][] = new Array(9).fill(0).map(() => new Array(9).fill(0))
    const columns: number[][] = new Array(9).fill(0).map(() => new Array(9).fill(0))
    const subboxes: number[][][] = new Array(3).fill(0).map(() => new Array(3).fill(0).map(() => new Array(9).fill(0)))

    for(let i = 0; i < 9; i++) {
        for (let j = 0; j < 9; j++) {
            const char = board[i][j]
            if (char !== '.') {
                // 这里减 1 是因为数独中的数字为 1 - 9，但是数组的下标从 0 开始
                const num = Number(char) - 1
                rows[i][num]++
                columns[j][num]++
                subboxes[Math.floor(i/3)][Math.floor(j/3)][num]++

                if (rows[i][num] > 1 || columns[j][num] > 1 || subboxes[Math.floor(i/3)][Math.floor(j/3)][num] > 1) {
                    return false
                }
            }
        }
    }

    return true
};