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LeetCode: Valid Sudoku 解题报告
Valid Sudoku
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character ‘.‘.
A partially filled sudoku which is valid.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
SOLUTION:
使用HashSet 行列，9块分别检查。
1 public class Solution { 2 public boolean isValidSudoku(char[][] board) { 3 if (board == null || board.length != 9 || 4 board[0].length != 9) { 5 return false; 6 } 7 8 HashSet<Character> set = new HashSet<Character>(); 9 10 // check the rows.11 for (int i = 0; i < 9; i++) {12 // clear the set at every row.13 set.clear();14 for (int j = 0; j < 9; j++) {15 if (!isValidChar(board[i][j], set)) {16 return false;17 }18 }19 }20 21 // check the columns.22 for (int i = 0; i < 9; i++) {23 // clear the set at every column.24 set.clear();25 for (int j = 0; j < 9; j++) {26 if (!isValidChar(board[j][i], set)) {27 return false;28 }29 }30 }31 32 // check the blocks.33 for (int i = 0; i < 9; i+=3) {34 for (int j = 0; j < 9; j+=3) {35 // clear the set at every block.36 set.clear();37 for (int k = 0; k < 9; k++) {38 if (!isValidChar(board[i + k / 3][j + k % 3], set)) {39 return false;40 } 41 }42 }43 }44 45 return true;46 }47 48 public boolean isValidChar(char c, HashSet<Character> set) {49 if (c == ‘.‘) {50 return true;51 }52 53 if (c < ‘0‘ || c > ‘9‘) {54 return false;55 }56 57 // Check if the character exit in the hashset.58 if (set.contains(c)) {59 return false;60 }61 62 set.add(c);63 64 return true;65 }66 }
主页君的GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/hash/IsValidSudoku.java
ref: http://www.ninechapter.com/solutions/valid-sudoku/
LeetCode: Valid Sudoku 解题报告
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