[leetcode-37]Sudoku Solver(java)

本文介绍了一种使用回溯算法解决数独问题的方法,通过深搜和剪枝技术,实现对空格的有效填充,确保唯一解的存在。

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问题描述:
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.
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这里写图片描述

分析:这道题使用的是回溯算法,回溯本质是深搜+剪枝。深搜又常常利用递归,然后当替换每个“.”时都判断是否有效。如果有效的话,就继续递归下去。

注意,一般递归函数都在开头位置判断是否结束,但是对于该问题而言,不大容易判断叶节点。所以这里采用的是利用返回值true或false来对树的深度进行控制。如果为solve到false时,就回溯。回溯的手段就是使用更改函数主体复位,并return。

代码如下:328ms

public class Solution {
    public void solveSudoku(char[][] board) {
        solve(board);
    }
    private boolean solve(char[][] board){
         for(int row = 0;row<9;row++){
             for(int col = 0;col<9;col++){
                 if(board[row][col] ==  '.'){
                     for(char i = '1';i<='9';i++){
                         board[row][col] = i;
                         if(isValid(board,row,col) && solve(board))
                            return true;
                         board[row][col] = '.';
                     }
                     return false;
                 }
             }
         }
         return true;
    }
    private boolean isValid(char[][] board,int row,int col){
        for(int i = 0;i<9;i++){
            if(i!=col && board[row][i] == board[row][col])
                return false;
        }
        for(int i = 0;i<9;i++){
            if(i!=row && board[i][col] == board[row][col])
                return false;
        }
        int beginRow = 3*(row/3);
        int beginCol = 3*(col/3);
        for(int i = beginRow;i<beginRow+3;i++){
            for(int j = beginCol;j<beginCol+3;j++){
                if(i!=row && j!=col && board[i][j] == board[row][col])
                    return false;
            }
        }
        return true;
    }
}
### 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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