【LeetCode】Sudoku Solver

Sudoku Solver 
Total Accepted: 20609 Total Submissions: 96910 My Submissions Question Solution 
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.
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【解題思路】
針對每個沒有數的點，對試1到9是否符合要求，如果是，回溯直到得到結果

Java AC

public class Solution {
    public int n, size;
	public ArrayList<Integer> list;
	public void solveSudoku(char[][] board) {
		n = board.length;
		list = new ArrayList<Integer>();
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (board[i][j] == '.') {
					list.add(i * n + j);
				}
			}
		}
		size = list.size();
		dfs(board, 0);
	}

	public boolean dfs(char[][] board, int cur) {
		if (cur == size) {
			return true;
		}
		int index = list.get(cur);
		int row = index / n;
		int col = index % n;
		for (int i = 1; i <= 9; i++) {
			if (isValidSudoku(board, row, col, (char)(i+'0'))) {
				board[row][col] = (char)(i+'0');
				if (dfs(board, cur+1)) {
					return true;
				}
				board[row][col] = '.';
			}
		}
		return false;
	}

	public boolean isValidSudoku(char[][] board, int i, int j, Character value) {
		for (int k = 0; k < n; k++) {
			if (k == j) {
				continue;
			}
			if (board[i][k] == value) {
				// System.out.println("1 "+i+" "+j );
				return false;
			}
		}
		for (int k = 0; k < n; k++) {
			if (k == i) {
				continue;
			}
			if (board[k][j] == value) {
				// System.out.println("2 "+i+" "+j );
				return false;
			}
		}
		int rowStart = i / 3 * 3;
		int rowEnd = rowStart + 3;
		int colStart = j / 3 * 3;
		int colEnd = colStart + 3;
		for (int k = rowStart; k < rowEnd; k++) {
			for (int l = colStart; l < colEnd; l++) {
				if (k == i && l == j) {
					continue;
				}
				if (board[k][l] == value) {
					// System.out.println("3 "+i+" "+j );
					return false;
				}
			}
		}
		return true;
	}
}

Python AC

class Solution:
    # @param board, a 9x9 2D array
    # Solve the Sudoku by modifying the input board in-place.
    # Do not return any value.
    def solveSudoku(self, board):
        num = []
        n = len(board)
        for i in xrange(n):
            for j in xrange(n):
                if board[i][j] == '.':
                    num.append(i * n + j)
        self.dfs(board, num, len(num), n, 0)
        print board

    def dfs(self, board, num, size, n, cur):
        if cur == size:
            return True
        index = num[cur]
        row = index / n
        col = index % n
        for i in xrange(1, 10):
            if self.isValidSudoku(board, row, col, chr(i+ord('0'))):
                temp = board[row]
                newTemp = ''
                for k in xrange(len(temp)):
                    if k == col:
                        newTemp += chr(i+ord('0'))
                    else:
                        newTemp += temp[k]
                board[row] = newTemp
                # board[row][col] = chr(i+ord('0'))
                if self.dfs(board, num, size, n, cur+1):
                    return True
                board[row] = temp
        return False

    def isValidSudoku(self, board, row, col, value):
        for i in xrange(9):
            if board[i][col] == value:
                return False
        for j in xrange(9):
            if board[row][j] == value:
                return False

        rowStart = row / 3 * 3
        rowEnd = rowStart + 3
        colStart = col / 3 * 3
        colEnd = colStart + 3
        for i in xrange(rowStart, rowEnd):
            for j in xrange(colStart, colEnd):
                if board[i][j] == '.':
                    continue
                if board[i][j] == value:
                    return False
        return True