leetcode037：Sudoku Solver

问题描述

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...

...and its solution numbers marked in red.

问题分析

数独解法基本靠暴力求解，在所有没有确定的位置对所有可能的解进行尝试，直接暴力解运行时间是153ms。所以在此之前先确定唯一解的位置，唯一解有两种类型。

1. 该位置在所在行、列、宫上都满足的情况下的候选集只有一个；
2. 该位置在所在行（列、宫）的所有未确定位置的候选集该值只出现一次。

代码

//Runtime: 17 msclass Solution {public:bool fun(vector<vector<char> > &board, int i, int j, int size) {for (; i < board.size(); i++){if (j == board[i].size()) j = 0;for (; j < board[i].size(); j++){if (board[i][j] == '.'){int a[10];int b[10];int c[10];//初始化memset(a, 0, sizeof(int) * 10);memset(b, 0, sizeof(int) * 10);memset(c, 0, sizeof(int) * 10);for (int m = 0; m < board[i].size(); m++){int k = board[i][m] - '0';if (k <= 9 && k >= 1){a[k] = 1;}}for (int m = 0; m < board.size(); m++){int k = board[m][j] - '0';if (k <= 9 && k >= 1){b[k] = 1;}}int s = i / 3 * 3;int t = j / 3 * 3;for (int m = 0; m < 9; m++){int k = board[s + m / 3][t + m % 3] - '0';if (k <= 9 && k >= 1){c[k] = 1;}}for (int k = 1; k <= 9; k++){if (!a[k] && (!b[k]) && (!c[k])){board[i][j] = k + '0';if (j + 1 == board[i].size() && fun(board, i + 1, 0, size)){return true;}else if (fun(board, i, j + 1, size)){return true;}board[i][j] = '.';}}return false;}}}return true;}void solveSudoku(vector<vector<char> > &board) {int a[9][10];//处理行int b[9][10];//处理列int c[9][10];//处理单元//初始化for (int i = 0; i < 9; i++){memset(a + i, 0, sizeof(int) * 10);memset(b + i, 0, sizeof(int) * 10);memset(c + i, 0, sizeof(int) * 10);}int m, n;int size = 0;for (int i = 0; i < board.size(); i++){for (int j = 0; j < board[i].size(); j++){int k = board[i][j] - '0';if (k <= 9 && k >= 1){size++;a[i][k] = 1;b[j][k] = 1;m = i / 3 * 3 + j / 3;c[m][k] = 1;}}}int x = 0;map<int, vector<int> > ma;map<int, vector<int>> mb[9];map<int, vector<int>> mc[9];int pre;do{pre = size;for (int i = 0; i < board.size(); i++){for (int j = 0; j < board[i].size(); j++){if (board[i][j] == '.'){int count = 0;m = i / 3 * 3 + j / 3;for (int k = 1; k <= 9; k++){if (!a[i][k] && (!b[j][k]) && (!c[m][k])){ma[k].push_back(j);mb[j][k].push_back(i * 9 + j);mc[m][k].push_back(i * 9 + j);}}}}map<int, vector<int> >::iterator it;for (it = ma.begin(); it != ma.end(); it++){if (it->second.size() == 1 && board[i][it->second.front()] == '.'){size++;board[i][it->second.front()] = '0' + it->first;a[i][it->first] = 1;b[it->second.front()][it->first] = 1;m = i / 3 * 3 + it->second.front() / 3;c[m][it->first] = 1;}}ma.clear();}for (int i = 0; i < 9; i++){for (auto it : mb[i]){int x = it.second.front() / 9;int y = it.second.front() % 9;if (it.second.size() == 1 && board[x][y] == '.'){size++;board[x][y] = '0' + it.first;a[x][it.first] = 1;b[y][it.first] = 1;m = x / 3 * 3 + y / 3;c[m][it.first] = 1;}}mb[i].clear();for (auto it : mc[i]){int x = it.second.front() / 9;int y = it.second.front() % 9;if (it.second.size() == 1 && board[x][y] == '.'){size++;board[x][y] = '0' + it.first;a[x][it.first] = 1;b[y][it.first] = 1;m = x / 3 * 3 + y / 3;c[m][it.first] = 1;}}mc[i].clear();}//cout << size << endl;//x++; } while (pre < size && size < 81);if (size == 81) return;fun(board, 0, 0, size);}};