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.

題目中說的,只要當前已經(jīng)填充的數(shù)字是合法的就可以,不一定要這個數(shù)獨是有解.（下面說的九宮格都是指3*3的網(wǎng)格）

因此只需要判斷9*9網(wǎng)格的每一行、每一列、9個小九宮格是否合法。即如果在每一行、每一列、每個9個小九宮格內(nèi)，某個數(shù)字重復(fù)出現(xiàn)了，當前數(shù)獨就是不合法的。

bool isValidSudoku(vector<vector<char> > &board) {// Note: The Solution object is instantiated only once.int row[9],col[9];for(int i = 0; i < 9; i++){memset(row,0,sizeof(int)*9);memset(col,0,sizeof(int)*9);for(int j = 0; j < 9; j++){if(board[i][j] != '.'){if(row[board[i][j]-'1'] > 0)return false;else row[board[i][j]-'1']++;}if(board[j][i] != '.'){if(col[board[j][i]-'1'] > 0)return false;else col[board[j][i]-'1']++;}}}for(int i = 0; i < 9; i+=3)for(int j = 0; j < 9; j+=3){memset(row,0,sizeof(int)*9);for(int a = 0; a < 3; a++)for(int b= 0; b < 3; b++)if(board[i+a][j+b] != '.'){if(row[board[i+a][j+b]-'1']>0)return false;else row[board[i+a][j+b]-'1']++;}}return true;}
public boolean isValidSudoku(char[][] board) {boolean[] visited = new boolean[10];//rowfor(int i=0; i<9; i++) {Arrays.fill(visited, false);for(int j=0; j<9; j++) {if( !check(visited, board[i][j]) ) {return false;}}}//colfor(int i=0; i<9; i++) {Arrays.fill(visited, false);for(int j=0; j<9; j++) {if( !check(visited, board[j][i]) ) {return false;}}}//sub matrixfor(int i=0; i<9; i+=3) {for(int j=0; j<9; j+=3) {Arrays.fill(visited, false);for(int k=0; k<9; k++) {if( !check(visited, board[i+k/3][j+k%3]) ) {return false;}}}}return true;}private boolean check(boolean[] visited, char c) {if(c == '.') {return true;}int num = c - '0';if(num<1 || num>9 ||visited[num]) {return false;} else {visited[num] = true;}return true;}

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