[LeetCode] 112: Sudoku Solver

最新推荐文章于 2019-08-23 22:27:00 发布

1.02^365的成长裂变

最新推荐文章于 2019-08-23 22:27:00 发布

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分类专栏： LeetCode2

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本文提供了一种使用深度优先搜索(DFS)算法解决数独问题的方法。通过预先填充已知数字并标记未解决的单元格，该算法能够递归地尝试每一个可能的数字直到找到解决方案。实现中还考虑了数独的唯一解假设。

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[Problem]

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.

[Solution]

class Solution {
public:
	/**
	 * DFS
	 */
	bool DFS(vector<vector<char> > &board, vector<pair<int, int> > unSolve, bool rowUsed[9][10], bool columnUsed[9][10], bool cubeUsed[9][10]){
		if(unSolve.size() == 0)return true;

		// the last one
		if(unSolve.size() == 1){
			int i = unSolve[0].first;
			int j = unSolve[0].second;

			// get the unused number in the row i
			int k = 1;
			for(k = 1; k <= 9; ++k){
				if(!rowUsed[i][k])break;
			}

			// there is no unused number in the row i, or the number is used in the column j or the cube (i/3)*3+j/3
			if(k == 10 || columnUsed[j][k] || cubeUsed[(i/3)*3+j/3][k])return false;

			// success
			board[i][j] = k + '0';
			return true;
		}
		else{
			int i = unSolve[0].first;
			int j = unSolve[0].second;

			// get the unused number in the row i
			for(int k = 1; k <= 9; ++k){
				// both the row i, the column j and the cube (i/3)*3+j/3 not use k
				if(rowUsed[i][k] || columnUsed[j][k] || cubeUsed[(i/3)*3+j/3][k])continue;

				// set the used
				rowUsed[i][k] = columnUsed[j][k] = cubeUsed[(i/3)*3+j/3][k] = true;
				
				// set the board
				board[i][j] = k + '0';

				// remove a pair from unSolve
				unSolve.erase(unSolve.begin());

				// solved
				if(DFS(board, unSolve, rowUsed, columnUsed, cubeUsed)){
					return true;
				}

				// unsolved, reset the marks
				rowUsed[i][k] = columnUsed[j][k] = cubeUsed[(i/3)*3+j/3][k] = false;
				board[i][j] = '.';
				unSolve.insert(unSolve.begin(), pair<int, int>(i, j));
			}
		}
		return false;
	}
	/**
	 * solve Sudoku
	 */
	void solveSudoku(vector<vector<char> > &board) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		// initial
		vector<pair<int, int> > unSolve;
		bool rowUsed[9][10], columnUsed[9][10], cubeUsed[9][10];	// 9 rows, 9 columns, 9 cubes
		memset(rowUsed, 0, sizeof(rowUsed));
		memset(columnUsed, 0, sizeof(columnUsed));
		memset(cubeUsed, 0, sizeof(cubeUsed));
		for(int i = 0; i < 9; ++i){
			for(int j = 0; j < 9; ++j){
				if(board[i][j] != '.'){
					rowUsed[i][board[i][j]-'0'] = true;
					columnUsed[j][board[i][j]-'0'] = true;
					cubeUsed[(i/3)*3+j/3][board[i][j]-'0'] = true;
				}
				else{
					unSolve.push_back(pair<int, int>(i, j));
				}
			}
		}

		// DFS
		DFS(board, unSolve, rowUsed, columnUsed, cubeUsed);
	}
};