LEETCODE 37. Sudoku Solver

本文介绍了一种使用递归回溯法解决数独问题的方法,该方法为LeetCode第36题的进阶版本。文章详细阐述了解题思路及其实现细节,包括如何避免无效的递归调用和如何正确地更新数独状态。

题目大意

是leetcode36的进阶版,要解出数独的答案
这里写图片描述

解题思路

就是用递归,无奈太渣,弄错了边界条件,而且思路还不够清晰,递归函数返回类型一开始写成void了,导致board里面的内容总是被清空,后来改成bool,并在清空内容之前做一下判断。

实现代码

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        bool rowFlag[9][9] = {false}, colFlag[9][9] = {false}, blockFlag[9][9] = {false};
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.') {
                    rowFlag[board[i][j] - '1'][i] = true;
                    colFlag[board[i][j] - '1'][j] = true;
                    blockFlag[board[i][j] - '1'][i / 3 * 3 + j / 3] = true;
                }
            }
        }
        recursion(board, rowFlag, colFlag, blockFlag, 0, 0);
    }
private:
    bool recursion(vector<vector<char>>& board, bool rowFlag[9][9], bool colFlag[9][9], bool blockFlag[9][9], int row, int col) {
        if (row == 9 && col == 0) {
            return true;
        }
        if (board[row][col] != '.') {
            col++;
            if (col > 8) {
                col = 0;
                row++;
            }
            return recursion(board, rowFlag, colFlag, blockFlag, row, col);
        } else {
            for (int i = 1; i <= 9; i++) {
                if (!rowFlag[i - 1][row] && !colFlag[i - 1][col] && !blockFlag[i - 1][row / 3 * 3 + col / 3]) {
                    rowFlag[i - 1][row] = true;
                    colFlag[i - 1][col] = true;
                    blockFlag[i - 1][row / 3 * 3 + col / 3] = true;
                    board[row][col] = i + '0';
                    int newRow = row, newCol = col + 1;
                    if (newCol > 8) {
                        newCol = 0;
                        newRow++;
                    }
                    bool flag = recursion(board, rowFlag, colFlag, blockFlag, newRow, newCol);
                    if (!flag) {
                        rowFlag[i - 1][row] = false;
                        colFlag[i - 1][col] = false;
                        blockFlag[i - 1][row / 3 * 3 + col / 3] = false;
                        board[row][col] = '.';
                    } else {
                        return true;
                    }
                }
            }
            return false;
        }
    }
};
### LeetCode Problem 37: Sudoku Solver #### Problem Description The task involves solving a partially filled Sudoku puzzle. The input is represented as a two-dimensional integer array `board` where each element can be either a digit from '1' to '9' or '.' indicating empty cells. #### Solution Approach To solve this problem, one approach uses backtracking combined with depth-first search (DFS). This method tries placing numbers between 1 and 9 into every cell that contains '.', checking whether it leads to a valid solution by ensuring no conflicts arise within rows, columns, and subgrids[^6]. ```cpp void solveSudoku(vector<vector<char>>& board) { backtrack(board); } bool backtrack(vector<vector<char>> &board){ for(int row = 0; row < 9; ++row){ for(int col = 0; col < 9; ++col){ if(board[row][col] != '.') continue; for(char num='1';num<='9';++num){ if(isValidPlacement(board,row,col,num)){ placeNumber(num,board,row,col); if(backtrack(board)) return true; removeNumber(num,board,row,col); } } return false; } } return true; } ``` In the provided code snippet: - A function named `solveSudoku()` initiates the process. - Within `backtrack()`, nested loops iterate over all positions in the grid looking for unassigned spots denoted by '.' - For any such spot found, attempts are made to insert digits ranging from '1' through '9'. - Before insertion, validation checks (`isValidPlacement`) ensure compliance with Sudoku rules regarding uniqueness per row/column/subgrid constraints. - If inserting a number results in reaching a dead end without finding a complete solution, removal occurs before trying another possibility. This algorithm continues until filling out the entire board correctly or exhausting possibilities when returning failure status upward along recursive calls stack frames. --related questions-- 1. How does constraint propagation improve efficiency while solving puzzles like Sudoku? 2. Can genetic algorithms provide alternative methods for tackling similar combinatorial problems effectively? 3. What optimizations could enhance performance further beyond basic DFS/backtracking techniques used here?
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