[LeetCode]SudokuSolver解决数独问题

一、 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.

A sudoku puzzle...

...and its solution numbers marked in red.

二、解决思路：

看起来数独是个比较麻烦的问题，实际上在用程序模拟的时候是个比较简单的过程，类似于八皇后问题，可采用回溯的思想解决。

这里主要创建了三个boolean的数组分别来记录棋盘board的行、列、块的状态，并用一个linkedlist记录board中为空的格子的位置，提高扫描效率。

注意每次回溯之后恢复这些数组和容器的状态即可。

三、源码：

import java.util.*;import java.io.*;public class SudokuSolver {public void solveSudoku(char[][] board) {        boolean[][] rowAppear = new boolean[9][9];        boolean[][] colAppear = new boolean[9][9];        boolean[][] blockAppear = new boolean[9][9];        List<Integer[]> list = new LinkedList<>();        for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {char tmp = board[i][j];if (tmp != '.') {rowAppear[i][tmp-'1']=true;colAppear[j][tmp-'1']=true;blockAppear[i/3*3+j/3][tmp-'1']=true;}else {list.add(new Integer[]{i,j});}}}        helper(board, rowAppear, colAppear, blockAppear, list);    }public boolean helper(char[][] board, boolean[][] rowAppear, boolean[][] colAppear, boolean[][] blockAppear, List<Integer[]> list) {if (list.isEmpty()) {return true;}Integer[] tmpPos = list.get(0);int x=tmpPos[0], y=tmpPos[1];list.remove(0);for (int i = 0; i < 9; i++) {if (!rowAppear[x][i] && !colAppear[y][i] && !blockAppear[x/3*3+y/3][i]) {board[x][y] = (char)('1'+i);rowAppear[x][i] = true;colAppear[y][i] = true;blockAppear[x/3*3+y/3][i] = true;if (helper(board, rowAppear, colAppear, blockAppear, list)) return true;board[x][y] = '.';rowAppear[x][i] = false;colAppear[y][i] = false;blockAppear[x/3*3+y/3][i] = false;}}list.add(0, tmpPos);return false;}public static void main(String[] args) {SudokuSolver sudokuSolver = new SudokuSolver();char[][] board = new char[9][9];Scanner scanner = new Scanner(System.in);for (int i = 0; i < 9; i++) {board[i] = scanner.next().toCharArray();}sudokuSolver.solveSudoku(board);for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {System.out.print(board[i][j]+" ");}System.out.print("\n");}}}