LeetCode [37. Sudoku Solver] 难度[hard]

题目

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

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

意思就是给出一个数独题，还没填数字的用'.'来表示。

算法思路：

数独游戏我们应该都玩过，规则就是每行没列每个九格宫都不能出现有相同的数字。那么在做这个题的时候，我们可以记录每一行、每一列、每个九宫格已经出现的数字，然后这样能就这可以知道每个格子可以尝试填哪些数字，填了数字以后立刻更新每一行、每一列、每个九宫格出现的数字，然后使用回溯便可以把题目做出来。该算法最坏情况下就是全部都没有填，相当于每行都做一个全排列，坏时间复杂度是O(9!^9)，算法实现如下：

class Solution {public:    bool fillSudoku(vector<vector<char>>& board, int i, int j){        if(i==9)    return true;        if(board[i][j]=='.'){            for(int k=1; k<=9; ++k){                if(!row[i][k]&&!col[j][k]&&!grid[i/3][j/3][k]){                    board[i][j] = '0'+k;                    row[i][k] = 1, col[j][k] = 1, grid[i/3][j/3][k] = 1;                    bool success = fillSudoku(board,i+(j+1)/9,(j+1)%9);                    if(success) return true;                    board[i][j] = '.';                    row[i][k] = 0, col[j][k] = 0, grid[i/3][j/3][k] = 0;                }            }        }        else            return  fillSudoku(board,i+(j+1)/9,(j+1)%9);                    return false;    }        void solveSudoku(vector<vector<char>>& board) {        memset(row,0,sizeof(row));        memset(col,0,sizeof(col));        memset(grid,0,sizeof(grid));        int n1 = board.size();        int n2 = board[0].size();        for(int i=0; i<n1; ++i){            for(int j=0; j<n2; ++j){                if(board[i][j]!='.'){                    row[i][board[i][j]-'0'] = 1;                    col[j][board[i][j]-'0'] = 1;                    grid[i/3][j/3][board[i][j]-'0'] = 1;                }            }        }        fillSudoku(board,0,0);        return ;    }private:    bool row[9][10],col[9][10],grid[3][3][10];};

可以直接在LeetCode上AC，耗时6ms