题目: 给定一个不完整的数独,要求填充好数独;最初给出的数独是有效的,且假设一定有答案;

举例:

A sudoku puzzle...

解题思路:

该题与青蛙走迷宫问题很相似,都是用深度优先;

代码如下:

 public class Solution {
public void solveSudoku(char[][] board) {
if(board == null || board.length < 9 || board[0].length < 9)
return;
solve(board, 0);
}
public boolean solve(char[][] board, int position)
{
if(position == 81) // position可以唯一确定一个坐标
return true;
int row = position / 9;
int col = position % 9;
if(board[row][col] == '.')
{
for(int i = 1; i <= 9; i++)
{
board[row][col] = (char)('0' + i);
if(checkValid(board, position)) // 检查将board[row][col]修改后,行列块是否有效
{
if(solve(board, position + 1)) // 深度搜索
return true;
}
board[row][col] = '.';
}
}
else
{
if(solve(board, position + 1))
return true;
}
return false;
}
public boolean checkValid(char[][] board, int position)
{
int row = position / 9;
int col = position % 9;
char target = board[row][col];
for(int j = 0; j < 9; j++)
{
if(j != col)
{
if(target == board[row][j]) // 判断除过col列,row行是否有taeget
return false;
}
if(j != row)
{
if(target == board[j][col]) // 判断除过row行,col列是否有target
return false;
}
}
int beginx = row / 3 * 3;
int beginy = col / 3 * 3;
for(int i = beginx; i < beginx + 3; i ++) // 块中是否有target
{
for(int j = beginy; j < beginy + 3; j ++)
{
if(i != row && j != col)
{
if(target == board[i][j])
return false;
}
}
}
return true; }
}

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