初级算法-10. 有效的数独

题目描述:

判断一个 9x9 的数独是否有效。只需要根据以下规则,验证已经填入的数字是否有效即可。

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

如图是一个部分填充的有效的数独

数独部分空格内已填入了数字,空白格用 '.' 表示。

示例 1:
输入:
[
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
输出: true
示例 2:
输入:
[
  ["8","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
输出: false

说明:

 
 
  • 一个有效的数独(部分已被填充)不一定是可解的。
  • 只需要根据以上规则,验证已经填入的数字是否有效即可。
  • 给定数独序列只包含数字 1-9 和字符 '.' 。
  • 给定数独永远是 9x9 形式的。
 

分析:1.可通过遍历每一列、每一行、每个3*3格子判断是否有重复元素

    此解法执行时间11ms ,超过82%Java提交记录(2019.4.27)

 1 class Solution {
 2     public static boolean isValidSudoku(char[][] board) {
 3         
 4         HashSet<Character> set=new HashSet<Character>();
       //判断每一行
5 for(int i=0;i<9;i++){ 6 for(int j=0;j<9;j++) { 7 if(board[i][j]!='.') { 8 if(!set.add(board[i][j])) 9 return false; 10 } 11 } 12 set.clear(); 13 }
      //判断每一列
14 for(int i=0;i<9;i++){ 15 for(int j=0;j<9;j++) { 16 if(board[j][i]!='.') { 17 if(!set.add(board[j][i])) 18 return false; 19 } 20 } 21 set.clear(); 22 }
      //判断每个3*3格子
23 for(int i=0;i<9;i+=3) { 24 for(int j=0;j<9;j+=3){ 25 for(int k=0;k<3;k++) 26 for(int l=0;l<3;l++) { 27 if(board[i+k][j+l]!='.') 28 if(!set.add(board[i+k][j+l])) 29 return false; 30 } 31 set.clear(); 32 } 33 34 } 35 return true; 36 } 37 }

2.提交记录中最快的:执行1ms 

 1 class Solution {
 2     boolean canPut(char[][] board, int i, int j, char v) {
 3         for(int k = 0; k < 9; k++) {
 4             if(board[i][k] == v 
 5                || board[k][j] == v 
 6                || board[i / 3 * 3 + k % 3][j / 3 * 3 + k / 3] == v) {
 7                 return false;
 8             }
 9         }
10         return true;
11     }
12     public boolean isValidSudoku(char[][] board) {
13         for(int i = 0; i < 9; i++) {
14             for(int j = 0; j <9; j++) {
15                 if(board[i][j] != '.') {
16                     char t = board[i][j];
17                     board[i][j] = '.';
18                     if(!canPut(board, i, j, t))
19                         return false;
20                     board[i][j] = t;
21                 }
22             }
23         }
24         return true;
25     }
26 }

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转载自www.cnblogs.com/hzhqiang/p/10781194.html