ValidSudoku

Change-Id: Ib833015581798bf95f546e0370340732a0e0cb12

Author: applewjg

ValidSudoku.java

@@ -0,0 +1,47 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 20, 2014
+ Problem:    Valid Sudoku
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/valid-sudoku/
+ Notes:
+ Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules (http://sudoku.com.au/TheRules.aspx).
+ The Sudoku board could be partially filled, where empty cells are filled with the character '.'.
+
+ Solution: 1. Traverse the Sudoku only once.
+           2. Bit manipulation. Use only one bit to represent a number. Space: sizeof(int) * (1+9+9).
+ */
+public class Solution {
+    public boolean isValidSudoku(char[][] board) {
+        boolean[] used = new boolean[9];
+        
+        for(int i=0;i<9;i++){
+            Arrays.fill(used,false);
+            for(int j = 0; j<9;j++){
+                if(check(board[i][j],used)==false) return false;
+            }
+            Arrays.fill(used,false);
+            for(int j = 0; j<9;j++){
+                if(check(board[j][i],used)==false) return false;
+            }
+        }
+        
+        for(int r = 0; r<3;r++){
+            for(int c = 0; c<3;c++){
+            Arrays.fill(used,false);
+                for(int i = r*3;i<r*3+3;i++){
+                    for(int j = c*3;j<c*3+3;j++){
+                        if(check(board[i][j],used)==false) return false;
+                    }
+                }
+            }
+        }
+        return true;
+    }
+    boolean check(char ch, boolean[] used){
+        if(ch=='.') return true;
+        if(used[ch-'1']) return false;
+        used[ch-'1']=true;
+        return true;
+    }
+}