題目鏈接
tag:

• Hard曹货；

question：
??Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

1. Each of the digits 1-9 must occur exactly once in each row.
2. Each of the digits 1-9 must occur exactly once in each column.
3. Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

A sudoku puzzle...

...and its solution numbers marked in red.

Note:

• The given board contain only digits 1-9 and the character '.'.
• You may assume that the given Sudoku puzzle will have a single unique solution.
• The given board size is always 9x9.

思路：
??這道求解數(shù)獨(dú)的題是在之前那道 Valid Sudoku 驗(yàn)證數(shù)獨(dú)的基礎(chǔ)上的延伸，之前那道題讓我們驗(yàn)證給定的數(shù)組是否為數(shù)獨(dú)數(shù)組，這道讓我們求解數(shù)獨(dú)數(shù)組灭将，跟此題類似的有Permutations 全排列，Combinations 組合項(xiàng)占遥， N-Queens N皇后問題等等，其中尤其是跟 N-Queens N皇后問題的解題思路及其相似痘煤，對于每個(gè)需要填數(shù)字的格子帶入1到9，每代入一個(gè)數(shù)字都判定其是否合法猿规，如果合法就繼續(xù)下一次遞歸衷快，結(jié)束時(shí)把數(shù)字設(shè)回'.'，判斷新加入的數(shù)字是否合法時(shí)姨俩，只需要判定當(dāng)前數(shù)字是否合法蘸拔，不需要判定這個(gè)數(shù)組是否為數(shù)獨(dú)數(shù)組，因?yàn)橹凹舆M(jìn)的數(shù)字都是合法的环葵，這樣可以使程序更加高效一些调窍，具體實(shí)現(xiàn)如代碼所示：

class Solution {
public:
    void solveSudoku(vector<vector<char> > &board) {
        if (board.empty() || board.size() != 9 || board[0].size() != 9) return;
        solveSudokuDFS(board, 0, 0);
    }
    bool solveSudokuDFS(vector<vector<char> > &board, int i, int j) {
        if (i == 9) return true;
        if (j >= 9) return solveSudokuDFS(board, i + 1, 0);
        if (board[i][j] == '.') {
            for (int k = 1; k <= 9; ++k) {
                board[i][j] = (char)(k + '0');
                if (isValid(board, i , j)) {
                    if (solveSudokuDFS(board, i, j + 1)) return true;
                }
                board[i][j] = '.';
            }
        } else {
            return solveSudokuDFS(board, i, j + 1);
        }
        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;
            }
        }
        return true;
    }
};

參考：http://www.cnblogs.com/grandyang/p/4421852.html