Add solution posts for leetcode 36 and 38.

Author: yanglr

docs/solutions/leetcode36 有效的数独[中等].md

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+---
+title: leetcode36 有效的数独【中等难度】|极客学长
+tags:
+  - leetcode
+  - 学习笔记
+  - 算法
+---
+
+### [36. 有效的数独](https://leetcode-cn.com/problems/valid-sudoku/)
+
+### 英文题目: Valid sudoku
+
+
+<table>    <tr>     <td bgcolor=white width=auto>&nbsp; ● &nbsp;难度: </td>    <td bgcolor=#F0AD4E width=auto><font color=white>中等</font></td>  </tr></table>
+
+
+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。
+
+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。（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：**
+
+![img](https://cdn.jsdelivr.net/gh/dbdgs/dbdgs.github.io/docs/.vuepress/public/img/sudu0.png)
+
+```
+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。
+```
+
+ 
+
+**提示：**
+
+- `board.length == 9`
+- `board[i].length == 9`
+- `board[i][j]` 是一位数字或者 `'.'`
+
+<br/>
+
+## 分析
+
+### 方法1、蛮力直接法
+
+使用set，
+对于行遍历: 每一行中, isValid: unique的数字数量+'.'的数量 = 9,
+对于列遍历：每一列中, isValid: unique的数字数量+'.'的数量 = 9,
+对于box遍历：每个3行3列九宫格中，isValid: unique的数字数量+'.'的数量 = 9。
+
+<br/>
+
+已AC代码:
+```cpp
+class Solution {
+public:
+    bool isValidSudoku(vector<vector<char>> &board)
+    {
+        bool isValid = true;
+
+        // 遍历行
+        for (int i = 0; i < 9; i++)
+        {
+            set<char> st;
+            vector<char> rowVec = board[i];
+            int dotCount = 0;
+            for (int k = 0; k < 9; k++)
+            {
+                if (rowVec[k] == '.')
+                {
+                    dotCount++;
+                }
+                else
+                    st.insert(rowVec[k]);
+            }
+            int uniqueCharCount = st.size();
+            if (uniqueCharCount + dotCount != 9)
+            {
+                isValid = false;
+            }
+        }
+
+        // 遍历列
+        for (int i = 0; i < 9; i++)
+        {
+            set<char> st;            
+            int dotCount = 0;
+            for (int k = 0; k < 9; k++)
+            {
+                if (board[k][i] == '.')
+                {
+                    dotCount++;
+                }
+                else
+                    st.insert(board[k][i]);
+            }
+            int uniqueCharCount = st.size();
+            if (uniqueCharCount + dotCount != 9)
+            {
+                isValid = false;
+            }
+        }
+
+        // 遍历小grid: 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
+        for (int si = 0; si <= 6; si += 3)                 
+            for (int sj = 0; sj <= 6; sj += 3)
+            {
+                set<char> st; 
+                int dotCount = 0;
+                for (int i = si; i < si + 3; i++)
+                {
+                    for (int j = sj; j < sj + 3; j++)
+                    {
+                        if (board[i][j] == '.')
+                            dotCount += 1;
+                        else
+                            st.insert(board[i][j]);
+                    }
+                }
+                if (st.size() + dotCount != 9)
+                    isValid = false;
+            }
+        return isValid;
+    }
+};
+```
+
+<br/>
+
+跟国外的小伙伴想到一块去了。
+<https://leetcode.com/problems/valid-sudoku/discuss/869625/easy-C%2B%2B-with-set>
+
+<br/>
+
+
+### 方法2：set插入方法 - 改进
+
+坐标中任意一点(i,j)，可以map到对应的的第几行第几列的方块(box)中，box的坐标为(i/3, j/3)。
+
+于是把一个小的九宫格中的数全压缩到一个box中，比如：
+
+![是否有插入失败的1](https://cdn.jsdelivr.net/gh/dbdgs/dbdgs.github.io/docs/.vuepress/public/img/sudu1.png)
+
+<br/>
+
+以最中间那个九宫格为例，使用int型的/3可以得到:
+
+![是否有插入失败的2](https://cdn.jsdelivr.net/gh/dbdgs/dbdgs.github.io/docs/.vuepress/public/img/sudu3.png)
+
+
+对于任意一个值不为'.'的字符，进行如下操作:
+
+1.把所在row的信息插入到大九宫格中;
+
+2.把所在column的信息插入到大九宫格中;
+
+3.把所在的小方块(box)的信息插入到大九宫格中。
+
+插入如果失败说明出现了重复。
+
+
+#### 已AC的C++代码:
+
+```cpp
+class Solution {
+public:
+    bool isValidSudoku(vector<vector<char>>& board) {
+        set<string> st;
+        for (int i = 0; i < 9; i++) {
+            for (int j = 0; j < 9; j++) {
+                char ch = board[i][j];
+                // 使用i / 3 + "," + j / 3 得到对应第几行第几列的方块(box)
+                if (ch != '.'){
+                    string val;
+                    val.push_back(ch);
+                    /* 对于任意一个值不为'.'的字符
+                       1.把所在row的信息插入到大九宫格中;
+                       2.把所在column的信息插入到大九宫格中;
+                       3.把所在的小方块(box)的信息插入到大九宫格中。
+                       插入如果失败说明出现了重复。 */
+                    if (!st.insert(val + " in row " + to_string(i)).second ||
+                        !st.insert(val + " in column " + to_string(j)).second ||
+                        !st.insert(val + " in box " + to_string(i / 3) + "," + to_string(j / 3)).second)
+                        return false;  /* set插入失败时，表示出现了重复 */
+                }
+            }
+        }
+        return true;        
+    }
+};
+```
+
+Java的HashSet有同样的写法，Java中插入失败，会出现 `set.Add() == false`。
+
+
+### 方法3：使用位操作
+
+此题，使用位操作，是几种解法中速度最快的算法了。
+
+具体做法是：
+
+![方法3](https://cdn.jsdelivr.net/gh/dbdgs/dbdgs.github.io/docs/.vuepress/public/img/sudu2.png)
+
+将大数独棋盘分成9个小棋盘，编号0~8。
+
+窗口中的每个小方格若有数字，必为 1 ~ 9 (记作k)，该方法适用于 遍历行/遍历列/遍历box。
+
+然后把 二进制数 1 左移 k 位，得到偏移量shift，后续使用按位或`|`来判断是否存在。
+
+
+#### 已AC的C++代码:
+
+```cpp
+class Solution {
+public:
+	bool isValidSudoku(vector<vector<char>>& board) {
+		vector<int> row(9); // row[j]表示第j 行的9个数字各自的存在情况，同理于col, boxes
+		vector<int> col(9);
+		vector<int> boxes(9);
+
+		int shiftInt = 0;
+		for (int i = 0; i < 9; i++)
+		{
+			for (int j = 0; j < 9; j++)
+			{
+				if (board[i][j] == '.')
+					continue;
+
+				shiftInt = 1 << (board[i][j] - '0');  // 转为二进制，移位结束后目标位为1，其他位均为0
+				/* 每个格子若有数字，必为 1 ~ 9，该方法适用于 遍历行/遍历列/遍历box  */
+				int boxPos = (i / 3) * 3 + j / 3; //将大数独棋盘分成9个小棋盘，编号0~8
+
+				// 如果当前数字shiftInt在row[j] 或col[i] 或 boxes中已经存在，&运算后不为0，
+				// 只有当前数字没出现过，&运算后为0
+				if ((col[i] & shiftInt) != 0 || (row[j] & shiftInt) != 0 || (boxes[boxPos] & shiftInt) != 0)
+					return false;
+
+				//第 n 位代表 n 这个数字是否存在(1→存在， 0→不存在)，同理于col[i]  boxes[boxPos]
+				row[j] |= shiftInt;
+				col[i] |= shiftInt;
+				boxes[boxPos] |= shiftInt;
+			}
+		}
+		return true;
+	}
+};
+```
+
+后两种方法，参考:
+
+<https://leetcode-cn.com/problems/valid-sudoku/solution/wei-yun-suan-qiu-jie-you-xiao-shu-du-c-b-sac7/>
+
+<https://www.youtube.com/watch?v=ceOLAY4XUOw&ab_channel=JacobHuang>