• 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
    };
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  • 原文地址:https://www.cnblogs.com/caoke/p/16734292.html
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