有效的数独

给定一个9*9数组

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。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

image

<small style="box-sizing: border-box; font-size: 12px;">上图是一个部分填充的有效的数独。</small>

数独部分空格内已填入了数字，空白格用 '.' 表示。

[
  ["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"]
]

说明:

• 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。
• 给定数独序列只包含数字 1-9 和字符 '.' 。
• 给定数独永远是 9x9 形式的。

我一开始的解法

var isValidSudoku = function(board) {
    let rowMapObj = {},columnMapObj = {},squareMapObj = {};
    for(let i=0;i<9;i++){
        for(let j=0;j<9;j++){
            let item = board[i][j];
            if(item === "."){
                continue;
            }
            //验证行
            if(!rowMapObj[i]) {
                rowMapObj[i] = []
            }
            if(rowMapObj[i].includes(item)){
                return false;
            }
            rowMapObj[i].push(item);
            //验证列
            if(!columnMapObj[j]) {
                columnMapObj[j] = []
            }
             if(columnMapObj[j].includes(item)){
                return false;
            }
            columnMapObj[j].push(item);
            //验证3 * 3数组
            let th = (i/3 | 0)+""+(j/3 | 0);
            if(!squareMapObj[th]){
              squareMapObj[th] = [];  
            }
            if(squareMapObj[th].includes(item)){
                return false;
            }
            squareMapObj[th].push(item)
        }
    }
    return true;
};

看了其他道友提交后总结的

var isValidSudoku = function(board) {
    let rowMapObj = Array(9)
          .fill(0)
          .map(() => ({})),
        columnMapObj = Array(9)
          .fill(0)
          .map(() => ({})),
        squareMapObj = Array(9)
          .fill(0)
          .map(() => ({}));
    
    for(let i=0;i<9;i++){
        
        for(let j=0;j<9;j++){
            let item = board[i][j];
            if(item !== "."){
                let squareIndex = (j/3 | 0) + (i/3|0)*3;
                if(squareMapObj[squareIndex][item] || rowMapObj[i][item] || columnMapObj[j][item] ){
                    return false;
                }
                squareMapObj[squareIndex][item] = rowMapObj[i][item] = columnMapObj[j][item] = true;
            }
        }
    }
    
    return true;
}

问题1：

如下写法麻烦吗？

 let rowMapObj = Array(9)
          .fill(0)
          .map(() => ({}))

问题2：

第二种解法在验证3*3宫内有无重复时，是映射为0-9的key，第一种解法是映射为"00"-"22"这样的key值，最核心的地方在于什么？还有其他映射方式吗？

我一开始也想映射为0-9这种，没想出来。

问题3:

能将 rowMapObj、columnMapObj 、squareMapObj 合并成一种数据结构吗？

问题4

第二种解法的let squareIndex = (j/3 | 0) + (i/3|0)*3; 为什么不是 let squareIndex = (i/3 | 0) + (j/3|0)*3;