力扣第37题 解数独 c++ 难~ 回溯

解数独 c++ 难~ 回溯"/>

力扣第37题 解数独 c++ 难~ 回溯

题目

37. 解数独

困难

相关标签

数组   哈希表   回溯   矩阵

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：

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。

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

思路和解题方法

1. 遍历行和列，找到一个空白格子（'.'）。
2. 从 1 到 9 依次尝试填入数字，判断当前数字是否合适放置在该格子位置。
3. 在 isValid 函数中，首先检查当前数字在所在行和列是否重复出现，如果重复则不合适。然后确定当前格子所在的 3x3 小方格起始位置，再检查该小方格内是否存在重复的数字。
4. 如果当前数字合适，则将其填入格子中，并继续递归地调用回溯函数 backtracking，以尝试解决剩下的空白格子。
5. 如果递归函数返回 true，说明成功找到了一组合适的数独解法，直接返回即可。
6. 如果递归函数返回 false，说明当前尝试的数字不合适，需要进行回溯操作，撤销当前数字的填入，并继续尝试下一个数字。
7. 如果遍历完所有的格子都没有返回 true，说明无法得到有效解，整个算法结束。

最终，当算法完成后，数独棋盘中的空白格子将被填满，或者数独问题没有解。

c++ 代码

class Solution {
private:bool backtracking(vector<vector<char>>& board) {for (int i = 0; i < board.size(); i++) {        // 遍历行for (int j = 0; j < board[0].size(); j++) { // 遍历列if (board[i][j] == '.') {for (char k = '1'; k <= '9'; k++) {     // (i, j) 这个位置放k是否合适if (isValid(i, j, k, board)) {board[i][j] = k;                // 放置kif (backtracking(board)) return true; // 如果找到合适一组立刻返回board[i][j] = '.';              // 回溯，撤销k}}return false;  // 9个数都试完了，都不行，那么就返回false}}}return true; // 遍历完没有返回false，说明找到了合适棋盘位置了}bool isValid(int row, int col, char val, vector<vector<char>>& board) {for (int i = 0; i < 9; i++) { // 判断行里是否重复if (board[row][i] == val) {return false;}}for (int j = 0; j < 9; j++) { // 判断列里是否重复if (board[j][col] == val) {return false;}}int startRow = (row / 3) * 3;int startCol = (col / 3) * 3;for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复for (int j = startCol; j < startCol + 3; j++) {if (board[i][j] == val ) {return false;}}}return true;}public:void solveSudoku(vector<vector<char>>& board) {backtracking(board);}
};

觉得有用的话可以点点赞，支持一下。

如果愿意的话关注一下。会对你有更多的帮助。

每天都会不定时更新哦  >人<  。