description:

數獨

Note:

Example:

Example 1:Input: [["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"] ] Output: trueExample 2:Input: [["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"] ] Output: false Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

answer:

class Solution { public:void solveSudoku(vector<vector<char>>& board) {if (board.empty() || board.size() != 9 || board[0].size() != 9) return;solveDFS(board, 0, 0); //從陣列的左上角開始逐行逐列的遞歸填入檢查}bool solveDFS(vector<vector<char>> &board, int i, int j) {if (i == 9) return true; // 如果行號到了9， 就證明已經全部遍歷完畢if (j >= 9) return solveDFS(board, i + 1, 0); // 如果列號大于等于9了，就證明這一行已經ok了，可以進行下一行的遞歸了。if (board[i][j] == '.') { // 如果是需要填入的空，那就從1到9挨個試，看看哪個填入是合法的for (int k = 1; k <= 9; k++){board[i][j] = (char)(k + '0');if (isValid(board, i, j)) { // 如果是合法的，就繼續遞歸，跳出遞歸之后就結束了也，這里必須要寫return ，這樣才能跳出循環，同時滿足函數返回值類型if (solveDFS(board, i, j + 1)) return true;}board[i][j] = '.'; // 這個是只有九個數都試過了但是都不行，就恢復原狀，然后就跳到最后那個 return false 上去了，這樣它前面一列那個數就相當于要換個k重新試試，這樣不斷往前改，總是能保證當前數之前填過的數都是正確的。}} else {return solveDFS(board, i, j + 1); // 如果不是需要去填的空，就跳過現在這個數繼續遞歸，這里一定要寫return}return false; }bool isValid(vector<vector<char>> &board, int i, int j) {for (int col = 0; col < 9; ++col) {if (col != j && board[i][j] == board[i][col]) return false;}for (int row = 0; row < 9; ++row) {if (row != i && board[i][j] == board[row][j]) return false;}for (int row = i/3*3; row < i/3*3 + 3; ++row) {for (int col = j/3*3; col < j/3*3 + 3; ++col) {if ((row != i || col != j) && board[i][j] == board[row][col]) return false; //這里是 || 只要行或者列任意一個數不一樣就要比一下}}return true;} };

relative point get√：

hint :

轉載于:https://www.cnblogs.com/forPrometheus-jun/p/11123306.html

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