标签:style blog http color io 2014 for re
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.
题解:递归。在每个空位上尝试放置0~9的数,然后递归的解决剩下的空位。
单独写一个判断board目前(x,y)处的数是否合法的函数 public boolean isValidSudoku(char[][] board,int x,int y) ,它就只用检查跟(x,y)同行,同列和同一个九宫格的元素是否和board[x][y]有重复即可。(其实在九宫格中只用判断4个和(x,y)不同行列的元素,因为和(x,y)同行列的我们已经判断过了)。
代码如下:
1 public class Solution { 2 public boolean isValidSudoku(char[][] board,int x,int y) { 3 //check for row x 4 for(int i = 0;i < 9;i++) 5 if(i!=y && board[x][i] == board[x][y]) 6 return false; 7 8 //check for column y 9 for(int i = 0;i < 9;i++) 10 if(i!= x &&board[i][y] == board[x][y]) 11 return false; 12 13 //check for the 3*3 square (x,y) belongs to 14 for(int i = 3 * (x/3);i<3*(x/3)+3;i++){ 15 for(int j = 3*(y/3);j<3*(y/3)+3;j++){ 16 if(i!=x && j != y && board[i][j] == board[x][y] ) 17 return false; 18 } 19 } 20 21 return true; 22 } 23 24 private boolean solveSudokuRecur(char[][] board){ 25 for(int i = 0;i < 9;i++){ 26 for(int j = 0;j < 9;j++){ 27 if(board[i][j] != ‘.‘) 28 continue; 29 for(int k = 1;k <= 9;k++){ 30 board[i][j] = (char)(k + ‘0‘); 31 if(isValidSudoku(board,i,j) && solveSudokuRecur(board)) 32 return true; 33 board[i][j] = ‘.‘; 34 } 35 return false; 36 } 37 } 38 return true; 39 } 40 public void solveSudoku(char[][] board) { 41 solveSudokuRecur(board); 42 } 43 }
注意之前做过的Valid Sudoku这道题是判断整个数独是否合法,而不是单独某个位置(x,y)是否合法,它需要遍历整个数独,所以这段代码不能拿来用了。
【leetcode刷题笔记】Sudoku Solver,布布扣,bubuko.com
标签:style blog http color io 2014 for re
原文地址:http://www.cnblogs.com/sunshineatnoon/p/3867695.html