Add another two C++ solutions for leetcode 36-数独.

Author: yanglr

cpp-leetcode/leetcode36-valid-sudoku_bit_operation.cpp

@@ -0,0 +1,61 @@
+#include <vector>
+#include <iostream>
+// #include <algorithm>
+using namespace std;
+
+class Solution {
+public:
+	bool isValidSudoku(vector<vector<char>>& board) {
+		vector<int> row(9); //row[j]表示第j 行的9个数字的存在情况，同理与col, boxes
+		vector<int> col(9);
+		vector<int> boxes(9);
+
+		int shiftCount = 0;
+		for (int i = 0; i < 9; i++)
+		{
+			for (int j = 0; j < 9; j++)
+			{
+				if (board[i][j] == '.')
+					continue;
+
+				shiftCount = 1 << (board[i][j] - '0');  // 转为二进制，目标位为1，其他位均为0
+				// 每个格子若有数字，必为 1 ~ 9，转换成对应的整型数字，1 → 0001，2 → 0010，3→0100，4→1000 ... 				
+				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shiftCount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] & shiftCount) != 0 || (row[j] & shiftCount) != 0 || (boxes[boxPos] & shiftCount) != 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] |= shiftCount;
+				col[i] |= shiftCount;
+				boxes[boxPos] |= shiftCount;
+			}
+		}
+		return true;
+	}
+};
+
+// Test
+int main()
+{
+    Solution sol;
+    vector<vector<char>> board  /* 不能使用=进行赋值初始化, 自动调用构造函数 */
+    {
+        {'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'}
+    };
+
+    bool isValid = sol.isValidSudoku(board);
+    cout << (isValid == true ? "true" : "false") << endl;
+
+    return 0;
+}

cpp-leetcode/leetcode36-valid-sudoku_direct_set.cpp

@@ -0,0 +1,56 @@
+#include <algorithm>
+#include <string>
+#include <vector>
+#include <set>       /* 使用set就够了，不一定需要unordered_set */
+#include <iostream>
+using namespace std;
+
+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;        
+    }
+};
+
+// Test
+int main()
+{
+    Solution sol;
+    vector<vector<char>> board  /* 不能使用=进行赋值初始化, 自动调用构造函数 */
+    {
+        {'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'}
+    };
+
+    bool isValid = sol.isValidSudoku(board);
+    cout << (isValid == true ? "true" : "false") << endl;
+
+    return 0;
+}