今天小编给大家带来力扣第37题-解数独，题目如下：

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

看到这题我的第一反应就是递归加回溯

1.为什么使用回溯

每个格子依次尝试一种可用的数字，在此基础上用同样的方式尝试填写下一格。若某个格子无数字可填，则回到前一格重新尝试下一个可用的数字。

状态记录：可以在一个格子盘上面直接填写数字操作，9*9二维数组中原地更新状态。

2.如何获取可用的数字

由限制条件入手：

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。

==>1-9遍历所有数字，同时满足上述限制条件的数字即为可选

3.如何记录限制条件

需要记录每行、每列、每3*3单元格数字的出现情况，用来获取可以的数字

==>使用数组来记录： rowArr[i][num]表示第i行有没有出现过num。

colArr[j][num]表示第j列有没有出现过num。

areaArr[k][num]表示第k个33单元格有没有出现过num。k = 3(row/3)+(col/3);

4.回溯过程

当前格子没有数字==>遍历并根据限制条件选取可靠数字填入，更新限制条件，递归填写下一格

当前格子已填入数字==>递归填写下一格

当前格子无法填写任何数字==>还原限制条件和已填入数字，回到上一格再尝试下一格可用数字

回溯状态还原时

5.递归停止条件

遍历到最后一格，让所有格子都填上了可用的数字

6.代码如下

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
 
var getAreaNumber = function(row,col)
{
	return Math.floor(row/3)*3 + Math.floor(col/3);
}  

var solveSuduku = function(board)
{
	let finish = false;
	let rowArr = new Array(9).fill(0).map(_=>new Array(10).fill(false));
	let colArr = new Array(9).fill(0).map(_=>new Array(10).fill(false));
	let areaArr = new Array(9).fill(0).map(_=>new Array(10).fill(false));
	for(let i=0;i<9;i++)
	{
		for(let j=0;j<9;j++)
		{
			let val = board[i][j];
			if(val!=".")
			{
				rowArr[i][val] = true;
				colArr[j][val] = true;
				areaArr[getAreaNumber(i,j)][val] = true; 
			}
		}
	}
	let dfs = function(curBoard,index)
	{
		if(index == 81)
		{
			finish = true;
			return;
		}
		let row = Math.floor(index/9);
		let col = Math.floor(index%9);
		let val = curBoard[row][col];
		if(val!=".")
		{
			dfs(curBoard,index+1);
		} 
		else
		{
			for(let num = 1;num<=9;num++)
			{
				let tempRow = rowArr[row][num];
				let tempCol = colArr[col][num];
				let  tempArea = areaArr[getAreaNumber(row,col)][num];
				if(!tempCol&&!tempRow&&!tempArea)
				{
					rowArr[row][num] = true;
					colArr[col][num] = true;
					areaArr[getAreaNumber(row,col)][num] = true;
					curBoard[row][col] = num.toString();
					dfs(curBoard,index+1);
					if(finish)
					{
						return;
					}
					curBoard[row][col] = ".";
					rowArr[row][num] = tempRow;
					colArr[col][num] = tempCol;
					areaArr[getAreaNumber(row,col)][num] = tempArea; 
				}
			}	 
		}
	} 
	dfs(board,0);
};