学习的进步-用马尔科夫链不停的寻找可能-37. 解数独

个人的大进步，将马尔科夫链的算法应用到了游戏中，有解决难题后的成就感

状态：通过

执行用时: 96 ms

内存消耗: 44.4 MB

提交时间：31 分钟前

 

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

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）
数独部分空格内已填入了数字，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：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：

 

 

提示：

board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据 保证 输入数独仅有一个解

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/sudoku-solver
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var solveSudoku = function(board) {
    //马尔科夫链不停的寻找可能
   
    // mayArr:空白点可能有哪些数字
    function getMayArr(x,y){
        const mayArr=['1','2','3','4','5','6','7','8','9']
        //比较行列，排查一些数字
        for(let i=0;i<9;i++){
            const d1=mayArr.indexOf(board[y][i])
            if(d1>-1){
                mayArr.splice(d1,1)
            }
            const d2=mayArr.indexOf(board[i][x])
            if(d2>-1){
                mayArr.splice(d2,1)
            }
        }
        const arr=[0,0,0,3,3,3,6,6,6];
        for(let i=arr[y];i<arr[y]+3;i++){
            for(let j=arr[x];j<arr[x]+3;j++){
                const d1=mayArr.indexOf(board[i][j])
                if(d1>-1){
                    mayArr.splice(d1,1)
                }
            }
        }
        return mayArr;
    }
    const data=[]
     //先统计空格的可能性
    for(let y=0;y<9;y++){
        for(let x=0;x<9;x++){
            if(board[y][x]==='.'){
                const arr=getMayArr(x,y)
                data.push([x,y,arr])
            }
        }
    }
    //data：这样从小到大排列可能性
    data.sort(function(p1,p2){
        return p1[2].length-p2[2].length
    })
    let preList=['']
    let list=[]
    const nArr=[0,0,0,3,3,3,6,6,6];
    for(let n=0;n<data.length;n++){
        //没有冲突，就继续增加链条
        const arr=data[n][2];
        arr.forEach(function(num){
            preList.forEach(function(str){
                //如果没有冲突
                let isHut=false;
                if(str.indexOf(num)>-1){
                      for(let i=0;i<str.length;i++){
                          if(str[i]===num){
                              if(data[i][0]===data[n][0]||data[i][1]===data[n][1]||nArr[data[i][0]]===nArr[data[n][0]]&&nArr[data[i][1]]===nArr[data[n][1]]){
                                  isHut=true;
                                  break;
                              }
                          }
                      }  
                }
                if(!isHut){
                    list.push(str+num)
                }
                
            })
        })
        preList=list
        list=[]
    }
    //preList 表示所有的可能
    if(preList.length>0){
        data.forEach(function(item,i){
            const x=item[0]
            const y=item[1]
            board[y][x]=preList[0][i]
        })
    }
    console.log(preList)
    return board
};