LeetCode刷题笔录Sudoku Solver

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.

和N-Queens那题想法差不多，对于每个格子，从1到9都填进去一遍试一下，然后check是否和当前已有的元素冲突。如果不冲突，则递归进行下一个格子。可以按行前进一格也可以按列前进一格。注意尝试填进去的数字再递归出来之后要改回'.‘。

public class Solution {
public void solveSudoku(char[][] board) {
solveSudoku(board, 0, 0);
}

public boolean solveSudoku(char[][] board, int row, int col){
if(col >= 9)
return solveSudoku(board, row + 1, 0);
if(row == 9)
return true;
if(board[row][col] == '.'){
for(char i = '1'; i <= '9'; i++){
board[row][col] = i;
if(isValid(board, row, col, i)){
if(solveSudoku(board, row, col + 1))
return true;
}
board[row][col] = '.';
}
}
else{
return solveSudoku(board, row, col + 1);
}
return false;
}
public boolean isValid(char[][] board, int row, int col, char value){
//check if duplicate exists in the same row or column
for(int i = 0; i < 9; i++){
if(i != col && value == board[row][i])
return false;
if(i != row && value == board[i][col])
return false;
}

//check if duplicate exists in the same cube
int cubeRowStart = row / 3 * 3;
int cubeColStart = col / 3 * 3;
for(int i = cubeRowStart; i < cubeRowStart + 3; i++){
for(int j = cubeColStart; j < cubeColStart + 3; j++){
if(i != row && value == board[i][j])
return false;
if(j != col && value == board[i][j])
return false;
}
}

return true;
}
}