leetCode 36.Valid Sudoku(有效的數獨) 解題思路和方法
阿新 • • 發佈:2019-02-06
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:
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.
思路:題目很簡單,主要是規則的理解,數獨的遊戲沒有玩過,不知道什麼規則,我以為任意9個方格1-9的個數都至多為1,誰知規則是特定的九個格內1-9的個數至多為1,其他不考慮。程式碼比較囉嗦,但思路清晰,如下:
public class Solution { //置為靜態變數 static Map<Character,Integer> map = new HashMap<Character,Integer>(); public boolean isValidSudoku(char[][] board) { //判斷每行 for(int i = 0; i < board.length; i++){ initMap();//每次均需初始化 for(int j = 0; j < board[0].length; j++){ //是數字 if(board[i][j] >= '0' && board[i][j] <= '9'){ if(map.get(board[i][j]) > 0){//說明重複數字 return false; }else{ map.put(board[i][j],1); } }else if(board[i][j] != '.'){//出現空格和0-9之外的字元 return false;//直接返回false } } } //判斷每列 for(int i = 0; i < board[0].length; i++){ initMap();//每次均需初始化 for(int j = 0; j < board.length; j++){ //是數字 if(board[j][i] >= '0' && board[j][i] <= '9'){ if(map.get(board[j][i]) > 0){//說明重複數字 return false; }else{ map.put(board[j][i],1); } }else if(board[j][i] != '.'){//出現空格和0-9之外的字元 return false;//直接返回false } } } //判斷九宮格 for(int i = 0; i < board.length - 2; i = i+3){//行{ for(int j = 0; j < board[0].length - 2; j=j+3){ initMap();//初始化 for(int m = i; m < i + 3;m++){ for(int n = j; n < j+3; n++){ //是數字 if(board[m][n] >= '0' && board[m][n] <= '9'){ if(map.get(board[m][n]) > 0){//說明重複數字 return false; }else{ map.put(board[m][n],1); } }else if(board[m][n] != '.'){//出現空格和0-9之外的字元 return false;//直接返回false } } } } } return true; } //初始化map為每個key均賦值0 private void initMap(){ for(char i = '0';i <= '9'; i++){ map.put(i,0); } } }