[LeetCode] 128: Valid Sudoku

最新推荐文章于 2020-01-30 17:22:51 发布

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本文介绍了一种验证数独是否有效的算法实现。该方法通过跟踪行、列和区块中出现过的数字来判断当前数独状态是否合法，而无需判断其是否可解。

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[Problem]

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

A partially filled sudoku which is valid.

[Analysis]
最开始以为需要判断是不是一个可解的数独，然后TLE。。。原来可以不用判断是否可解，只需要判断当前数独状态是否合法即可。

[Solution]

class Solution {
public:
	bool isValidSudoku(vector<vector<char> > &board) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		
		// initial
		bool rowUsed[9][10], columnUsed[9][10], cubeUsed[9][10];
		memset(rowUsed, 0, sizeof(rowUsed));
		memset(columnUsed, 0, sizeof(columnUsed));
		memset(cubeUsed, 0, sizeof(cubeUsed));

		// set rowUsed, columnUsed, cubeUsed
		vector<pair<int, int> > unSolved;
		for(int i = 0; i < 9; ++i){
			for(int j = 0; j < 9; ++j){
				// filled
				if(board[i][j] != '.'){
					// set rowUsed
					if(rowUsed[i][board[i][j]-'0'])return false;
					rowUsed[i][board[i][j]-'0'] = true;

					// set columnUsed
					if(columnUsed[j][board[i][j]-'0'])return false;
					columnUsed[j][board[i][j]-'0'] = true;

					// set cubeUsed
					if(cubeUsed[i/3*3+j/3][board[i][j]-'0'])return false;
					cubeUsed[i/3*3+j/3][board[i][j]-'0'] = true;
				}
				// empty
				else{
					unSolved.push_back(pair<int, int>(i, j));
				}
			}
		}

		return true;
	}
};