leetcode 37 Sudoku Solver

本文详细阐述了使用回溯法解决数独问题的算法实现过程,包括判断安全性和递归搜索步骤,确保数独解决方案的正确性。

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class Solution {
public:
    bool safe(vector<vector<char>>&board,int i,int j,int num)
    {
        for(int k=0;k<9;++k)
        {
            if(board[i][k]==num+'0' || board[k][j]==num+'0')return false;
        }
        int row = i/3;
        int col = j/3;
        
        for(int a=0;a<3;++a){
            for(int b=0;b<3;++b){
                if(board[3*row+a][3*col+b]==num+'0')return false;
            }
        }
        
        return true;
    }
    
    bool Is_true(vector<vector<char>>&board,int row,int col)
    {
        if(col > 8){col %= 9;row++;}
        if(row > 8)return true;
        
        if(board[row][col] == '.'){
            for(int i=1;i<10;++i){
                if(safe(board,row,col,i)){
                    board[row][col]=i+'0';
                    if(Is_true(board,row,col+1))return true;//要记住找所有解和找一个解的区别!!!否则会回溯回去导致前面找到的解又不见了
                    board[row][col]='.';
                }
            }
        }
        
        else{
            if(Is_true(board,row,col+1))return true;
        }
        
        return false;
    }
    
    void solveSudoku(vector<vector<char>>& board) {
       
        Is_true(board,0,0);
    }
};

### 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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