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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:
1 public class Solution { 2 public void solveSudoku(char[][] board) { 3 if(board == null || board.length < 9 || board[0].length < 9) 4 return; 5 solve(board, 0); 6 } 7 public boolean solve(char[][] board, int position) 8 { 9 if(position == 81) // position可以唯一确定一个坐标 10 return true; 11 int row = position / 9; 12 int col = position % 9; 13 if(board[row][col] == ‘.‘) 14 { 15 for(int i = 1; i <= 9; i++) 16 { 17 board[row][col] = (char)(‘0‘ + i); 18 if(checkValid(board, position)) // 检查将board[row][col]修改后,行列块是否有效 19 { 20 if(solve(board, position + 1)) // 深度搜索 21 return true; 22 } 23 board[row][col] = ‘.‘; 24 } 25 } 26 else 27 { 28 if(solve(board, position + 1)) 29 return true; 30 } 31 return false; 32 } 33 public boolean checkValid(char[][] board, int position) 34 { 35 int row = position / 9; 36 int col = position % 9; 37 char target = board[row][col]; 38 for(int j = 0; j < 9; j++) 39 { 40 if(j != col) 41 { 42 if(target == board[row][j]) // 判断除过col列,row行是否有taeget 43 return false; 44 } 45 if(j != row) 46 { 47 if(target == board[j][col]) // 判断除过row行,col列是否有target 48 return false; 49 } 50 } 51 int beginx = row / 3 * 3; 52 int beginy = col / 3 * 3; 53 for(int i = beginx; i < beginx + 3; i ++) // 块中是否有target 54 { 55 for(int j = beginy; j < beginy + 3; j ++) 56 { 57 if(i != row && j != col) 58 { 59 if(target == board[i][j]) 60 return false; 61 } 62 } 63 } 64 return true; 65 66 } 67 }
Leetcode37--->Sudoku Solver(填充数独)
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原文地址:http://www.cnblogs.com/leavescy/p/5899277.html