leetcode 37. Sudoku Solver

本文介绍了一种使用递归回溯法解决数独谜题的方法。通过检查每一格的数值合法性来逐步填充空格,实现数独谜题的解答。

1.题目

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.


A sudoku puzzle...

求解数独。。。

2.思路 

利用递归,先放一个数在格子上,如果合格,再放下一个格子,如果不合格,回溯。

class Solution {
public:
    bool isValid(vector<vector<char>>& board,int position) { //测试当前position位置上的数是否合法
    int row = position/9;
    int col = position%9;
    int gid = (row/3)*3 + col/3; //第(row/3*3 + col/3)个小方块的起始位置

    for(int i=0;i<9;i++)
    {
        //check row
        if(i != col && board[row][i] == board[row][col])
            return false;
        
        //check col
        if(i != row && board[i][col] == board[row][col])
            return false;
        
        //check subgrid
        int r = gid/3*3+i/3;
        int c = gid%3*3+i%3;
        if((r != row || c != col) && board[r][c] == board[row][col])
            return false;
    }
    return true;
}
bool solve(vector<vector<char>>& board,int position){
    if(position == 81)
        return true;
    
    int row = position/9;
    int col = position%9;
    
    if(board[row][col] == '.')
    {
        for(int k=1;k<=9;k++)
        {
            board[row][col] = k + '0';
            if(isValid(board,position))
                if(solve(board,position+1))
                    return true;
        }
        board[row][col] = '.';
        return false;
    }
    else return  solve(board,position+1);
}

void solveSudoku(vector<vector<char>>& board) {
    solve(board,0);
}
};


### 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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