【LeetCode】109.Sudoku Solver

题目描述(Hard)

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.

题目链接

https://leetcode.com/problems/sudoku-solver/description/

算法分析

深搜，对每个位置判断，时间复杂度，可与【LeetCode】14.Valid Sudoku一起食用。

提交代码：

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        DFSsolveSudoku(board);
    }
    
private:
    bool DFSsolveSudoku(vector<vector<char>>& board) {
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') {
                    for (int num = 0; num < 9; ++num) {
                        board[i][j] = '1' + num;
                        if (isValid(board, i, j) && DFSsolveSudoku(board))
                            return true;
                        board[i][j] = '.';
                    }
                    return false;
                }
            }
        }
        return true;
    }
    
    bool isValid(vector<vector<char>>& board, int x, int y) {
        // 所在列有效性检测
        for (int i = 0; i < 9; ++i) {
            if (i != x && board[i][y] == board[x][y])
                return false;
        }
        
        // 所在行有效性检测
        for (int j = 0; j < 9; ++j) {
            if (j != y && board[x][j] == board[x][y])
                return false;            
        }
        
        // 所在子格有效性检测
        for (int i = 3* (x / 3); i < 3* (x / 3 + 1); ++i)
            for (int j = 3* (y / 3); j < 3* (y / 3 + 1); ++j)
                if((i != x || j != y) && board[i][j] == board[x][y])
                    return false;

        return true;
    }
};