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这个系列没啥花头，就是纯 leetcode 题目拆解分析，不求用骚气的一行或者小众取巧解法，而是用清晰的代码和足够简单的思路帮你理清题意。让你在面试中再也不怕算法笔试。

120. 有效的数独 (valid-sudoku)

题目

leetcode 传送门

这里不贴题了，leetcode打开就行，题目大意：

请你判断一个 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。

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。

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。

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。

基本思路

判断是否合法其实比较简单，就是用横，纵，子盒子(3x3)设置3个数组，然后依次判断每个格子，这个格子里的数如果跟已知当前行列或子盒子的数重复，就说明不合法，否则就是合法，并在这三个数组中推入，表示该位置填上当前数。

行列，用 i, j 直接判断行列即可，子盒子的 index 判断稍微复杂些，看这个图

box_index = (i / 3) * 3 + j / 3，其中 / 是整数除法

知道 (i, j) 我们就能知道是哪个子盒子，也就是根据坐标，获取所在的子盒子的索引

写法实现

const isValidSudoku = (board) => {
  // 横，纵，小盒子的初始化，表示已经填上的数字，开始都是空的
  let rowsUsed = new Array(9).fill(0).map(it => [])
  let colsUsed = new Array(9).fill(0).map(it => [])
  let boxesUsed = new Array(9).fill(0).map(it => [])

  for (let i = 0; i < 9; i++) {
    for (let j = 0; j < 9; j++) {

      const cur = board[i][j];

      // 空的无所谓，跳过，因为我们判断的是现有填上的是否合法
      if (cur !== '.') {

        // 当前横排已经有此元素，直接返回 false
        if (rowsUsed[i].includes(cur)) {
          return false
        } else {
          // 否则就填上，然后推入该已用数组
          rowsUsed[i].push(cur)
        }
        
        // 竖排相同道理
        if (colsUsed[j].includes(cur)) {
          return false;
        } else {
          colsUsed[j].push(cur);
        }

        // 9个小方格稍微复杂点，算出大方块的 index，然后还是相同的判断9个数是否被用过
        const boxIndex = Math.floor(i / 3) * 3 + Math.floor(j / 3);

        if (boxesUsed[boxIndex].includes(cur)) {
          return false;
        } else {
          boxesUsed[boxIndex].push(cur);
        }
      }

    }
  }

  return true;
};

console.log(isValidSudoku(
[[".","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"]]
));

121. 解数独 (sudoku-solver)

题目

leetcode 传送门

这里不贴题了，leetcode打开就行，题目大意：

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：

基本思路

如果对 DFS + 回溯问题不熟悉请先看看之前这个系列的问题

另外数独和之前的文章 N皇后有点类似，可以先看看

这种题目基本都可以用暴力法求解，就是穷举，但是我们应该多思考更好的解决方案，如果你是个玩数独的高手，或者简单来说正常思维，我们都会在数独中，先找那种好填的填写，什么是好填? 如果一行中，全都已经写好就差一个数没填，那是不是只能填那个数，你只有一种填法，只需要枚举一个数就行了，在深度遍历的路上相当于就一条道，简单吧。当然现实不可能这么简单，但思想就是，先从选择少的那些路开始走,这能大大减少 DFS 的复杂度，剪去了许多无用枝条。

我们也可以做一次预处理，先把每个空格子能填的数都给他算好，然后同样 DFS 先遍历少的，然后回溯，最终找到一个正确解。详细做法根据我的代码注释可以清楚地理解，代码虽多，但我们的思路一定要清晰，不要畏惧，看懂了很简单。

写法实现

const solveSudoku = (board) => {
  // 枚举值
  const enumVal = ['1', '2', '3', '4', '5', '6', '7', '8', '9']; 

  const rows = new Array(9).fill(0).map(it => new Set(enumVal));    // 每一行的 可选数集
  const cols = new Array(9).fill(0).map(it => new Set(enumVal));    // 每一列的 可选数集
  const blocks = new Array(9).fill(0).map(it => new Set(enumVal));  // 每一子盒的 可选数集
  
  // 根据坐标，获取所在的子盒子的索引
  const getBoxIndex = (i, j) => {
    return Math.floor(i / 3) * 3 + Math.floor(j / 3);
  };

  // 根据现有的已填的数字，精简可选数据集，把不能填写的数字从可选数据集中删除
  for (let i = 0; i < 9; i++) {    
    for (let j = 0; j < 9; j++) {
      if (board[i][j] !== ".") {
        // 当前行出现过这个数字，这个数字就不能在这一行出现，就应该从这一行的`可选`数据集中删除
        // 可选的意思就是 选了之后仍然合法
        rows[i].delete(board[i][j]);
        cols[j].delete(board[i][j]);
        blocks[getBoxIndex(i, j)].delete(board[i][j]);
      }
    }
  }

  const dfs = (i, j) => {
    // 列超过最边界坐标
    if (j > 8) {
      // 行+1
      i++;
      // 列回到起点
      j = 0;
      if (i > 8) {
        // 都填完了
        return true;
      }
    }

    // 如果不是空白格，递归填下一格
    if (board[i][j] !== ".") {
      return dfs(i, j + 1)
    } 

    const boxIndex = getBoxIndex(i, j); // 获取所在小框的索引

    // 枚举出所有选择：1-9
    for (let num = 1; num <= 9; num++) { 
      const cur = String(num);
      // 当前选择必须在三个set中都存在，如果有一个不存在，就说明发生了冲突，跳过该选择，这就是剪枝
      if (!rows[i].has(cur) || !cols[j].has(cur) || !blocks[boxIndex].has(cur)) continue;

      // 作出选择，填上该数
      board[i][j] = cur;
      // 删掉这个可填选项
      rows[i].delete(cur);  
      cols[j].delete(cur);
      blocks[boxIndex].delete(cur);

      if (dfs(i, j + 1)) return true; // 如果基于当前选择，填下一个，最后可解出数独，直接返回真
      
      // 回溯，恢复为空白格
      board[i][j] = ".";
      // 将之前删掉的当前数字，加回来
      rows[i].add(cur);     
      cols[j].add(cur);
      blocks[boxIndex].add(cur);
    }

    // 尝试了1-9，每个都往下递归，都不能做完，返回false
    return false;  
  };

  // 填格子的起点
  dfs(0, 0);
  return board;
};

let 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"]]

console.log(solveSudoku(board))

另外向大家着重推荐下这个系列的文章，非常深入浅出，对前端进阶的同学非常有作用，墙裂推荐！！！核心概念和算法拆解系列

今天就到这儿，想跟我一起刷题的小伙伴可以加我微信哦点击此处交个朋友 Or 搜索我的微信号infinity_9368，可以聊天说地加我暗号 "天王盖地虎" 下一句的英文，验证消息请发给我 presious tower shock the rever monster，我看到就通过，加了之后我会尽我所能帮你，但是注意提问方式，建议先看这篇文章：提问的智慧

前端算法必刷题系列[63]

120. 有效的数独 (valid-sudoku)

标签

题目

基本思路

写法实现

121. 解数独 (sudoku-solver)

标签

题目

基本思路

写法实现

参考