前端算法题解有效的数独

时间:2022-09-29 前端_奔跑的蜗牛人气:0

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 '.'

解题思路-分别处理

分别处理每一行、每一列以及每一个九宫格，哪一部分验证失败，都返回 false，如果都验证通过，返回 true。

代码实现

function getOrigin(){
    return {
        '1':0,
        '2':0,
        '3':0,
        '4':0,
        '5':0,
        '6':0,
        '7':0,
        '8':0,
        '9':0,
    }
}
function checkArr(arr){
    const counts = getOrigin()
    for(let i = 0;i<9;i++){
        if(counts[arr[i]]){
            return false
        }else{
            counts[arr[i]]++
        }
    }
    return true
}
var isValidSudoku = function(board) {
    // 处理每一行
    for(let i = 0;i<9;i++){
        if(!checkArr(board[i])){
            return false
        }
    }
    // 处理每一列
    for(let i = 0;i<9;i++){
        const arr = []
        for(let j = 0;j<9;j++){
            if(board[j][i] === '.'){
                continue
            }
            arr.push(board[j][i])
        }
        if(!checkArr(arr)){
            return false
        }
    }
    // 处理 9 个九宫格
    for(let i = 0;i<3;i++){
        for(let j = 0;j<3;j++){
            const arr = []
            for(let k = j*3;k<j*3+3;k++){
                for(let h = 3*i;h<3*i+3;h++){
                    if(board[k][h] === '.'){
                        continue
                    }
                    arr.push(board[k][h])
                }
            }
            if(!checkArr(arr)){
                return false
            }
        }
    }
    return true
}

解题思路-一次扫描判断所有

首先创建 lines 记录每一行出现的数字的次数，columns 记录每一列出现的数字的次数，scratchableLatexs 记录每一个九空格出现的数字的次数。

然后双层循环可以遍历输入数组中的每一个元素，根据当前 i,j 值判断属于哪一行，哪一列以及哪一个九宫格，分别判断即可。

代码实现

function getOrigin(){
    return {
        '1':0,
        '2':0,
        '3':0,
        '4':0,
        '5':0,
        '6':0,
        '7':0,
        '8':0,
        '9':0,
    }
}
var isValidSudoku = function(board) {
    const lines = []
    const columns = []
    const scratchableLatexs = []
    for(let i = 0;i<9;i++){
        lines[i] = getOrigin()
        columns[i] = getOrigin()
        scratchableLatexs[i] = getOrigin()
    }
    for(let i = 0;i<9;i++){
        for(let j = 0;j<9;j++){
            const item = board[i][j]
            if(item === '.'){
                continue
            }
            if(lines[i][item]){
                return false
            }else{
                lines[i][item]++
            }
            if(columns[j][item]){
                return false
            }else{
                columns[j][item]++
            }
            if(i<3){
                if(j<3){
                    if(scratchableLatexs[0][item]){
                        return false
                    }else{
                        scratchableLatexs[0][item]++
                    }
                }else if(j<6){
                    if(scratchableLatexs[1][item]){
                        return false
                    }else{
                        scratchableLatexs[1][item]++
                    }
                }else{
                    if(scratchableLatexs[2][item]){
                        return false
                    }else{
                        scratchableLatexs[2][item]++
                    }
                }
            }else if(i<6){
                if(j<3){
                    if(scratchableLatexs[3][item]){
                        return false
                    }else{
                        scratchableLatexs[3][item]++
                    }
                }else if(j<6){
                    if(scratchableLatexs[4][item]){
                        return false
                    }else{
                        scratchableLatexs[4][item]++
                    }
                }else{
                    if(scratchableLatexs[5][item]){
                        return false
                    }else{
                        scratchableLatexs[5][item]++
                    }
                }
            }else{
                if(j<3){
                    if(scratchableLatexs[6][item]){
                        return false
                    }else{
                        scratchableLatexs[6][item]++
                    }
                }else if(j<6){
                    if(scratchableLatexs[7][item]){
                        return false
                    }else{
                        scratchableLatexs[7][item]++
                    }
                }else{
                    if(scratchableLatexs[8][item]){
                        return false
                    }else{
                        scratchableLatexs[8][item]++
                    }
                }
            }
        }
    }
    return true
}

至此我们就完成了 leetcode-36-有效的数独，更多关于前端算法题解有效的数独的资料请关注其它相关文章！

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