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14解数独回溯

news 2026/5/24 11:47:41

37. 解数独

困难

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提示

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

class Solution {
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kbool result = backtracking(board);//只有找到正确解才往上返回，若是false，那么回溯撤销原数字之后继续递归if(result) return true; //找到一组解就一直往上返回直到第一层// if (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);}
};

this version only 需要 find一组解就可以直接返回了
第一层递归中（包括后面除了最后一层的递归也是一样）返回true不会在最后末尾那一行返回，而是在for循环中，意味着我找到一个空格，本层填1~9其中一个数，最终找到了题解（正确的数独），最下面那一行true只在最后一个数都已经填进去了（且全部合法），下一层没有空格，才会执行最后一行的return true，某一层返回false的情况是我这个方格是空格，但是填1~9都无法凑出合法的数独，所以返回false之后让他的上一层甚至是更前面的层重新选一个数填进

class Solution {// 保存所有解（和N皇后格式完全一样！）vector<vector<string>> result;// 改成void，找所有解void backtracking(vector<string>& board) {for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {if (board[i][j] == '.') {for (char k = '1'; k <= '9'; k++) {if (isValid(i, j, k, board)) {board[i][j] = k;backtracking(board);  // 不return，继续找board[i][j] = '.';    // 回溯}}return; // 没数字可填，回溯让前面层填的数字换一换}}}// ✅ 找到一组解，保存！result.push_back(board);}// 判断合法（不变）bool isValid(int row, int col, char val, vector<string>& board) {// 检查行for (int i = 0; i < 9; i++) {if (board[row][i] == val) return false;}// 检查列for (int i = 0; i < 9; i++) {if (board[i][col] == val) return false;}// 检查3x3九宫格int sr = (row / 3) * 3;int sc = (col / 3) * 3;for (int i = sr; i < sr + 3; i++) {for (int j = sc; j < sc + 3; j++) {if (board[i][j] == val) return false;}}return true;}public:vector<vector<string>> solveSudoku(vector<vector<char>>& board) {result.clear();// 把输入的 vector<vector<char>> 转换成 vector<string>vector<string> strBoard;for (auto& row : board) {strBoard.push_back(string(row.begin(), row.end()));}backtracking(strBoard);return result;}
};