[leetcode]Sudoku Solver

新博文地址：[leetcode]Sudoku Solver

 

Sudoku Solver

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.

 好了，最后一道题了，一直对数独特别害怕，不知道为啥，但是仔细看了这道题，貌似没啥难度，跟八皇后问题一样，典型的回溯算法，当然也可以用迭代实现，我同学用了迭代实现的，因此我就写了一个递归版本的，算法思想就是典型的DFS，因此这里不再啰嗦。

尴尬的是，DFS在找到一个结果的时候会继续回溯寻找下一个可能的结果，因此我又设置了一个全局变量来添加控制。这无疑让代码可读性变差，如果谁有更好的方法：找到一个解之后立马跳出整个递归过程，请告之

代码如下：

        boolean over = false;
	public void solveSudoku(char[][] board) {
		if (board == null || board.length == 0
				|| board.length != board[0].length || board.length % 9 != 0) {
			return;
		}
		int height = board.length;
		int width = board[0].length;
		dfs(0, 0, board, height, width);
	}

	private void dfs(int row, int column, char[][] board, int height, int width) {
		if(over) return;
		if (row == board.length ) {
			over = true;
			return;
		}
		if (board[row][column] != '.' ) {
			if (column < width - 1 && !over) {
				dfs(row, column + 1, board, height, width);
			} else if (column == width - 1 && !over) {
				dfs(row + 1, 0, board, height, width);
			}
		} else {
			for (char c = '1'; c <= '9' && !over; c++) {
				if (!isConflict(c, row, column, board)) {
					board[row][column] = c;
					if (column < width - 1) {
						dfs(row, column + 1, board, height, width);
					} else if (column == width - 1) {
						dfs(row + 1, 0, board, height, width);
					}
				}
			}
			if(!over)board[row][column] = '.';
		}
	}

	private boolean isConflict(char num, int row, int column, char[][] board) {
		int height = board.length;
		int width = board[0].length;
		for (int i = 0; i < width; i++) {
			if (i != column && board[row][i] == num) {
				return true;
			}
		}
		for (int i = 0; i < height; i++) {
			if (i != row && board[i][column] == num) {
				return true;
			}
		}
		int beginRow = (row / 3) * 3;
		int endRow = beginRow + 2;
		int beginColumn = (column / 3) * 3;
		int endColumn = beginColumn + 2;
		for (int i = beginRow; i <= endRow; i++) {
			for (int j = beginColumn; j <= endColumn; j++) {
				if (i == row && j == column)
					continue;
				if (board[i][j] == num)
					return true;
			}
		}
		return false;
	}