Leetcode 37. Sudoku Solver

https://leetcode.com/problems/sudoku-solver/

DFS is often the right solution for this kind of problems. Try every space and judge if it is valid.

Solution:

# T:O((9!)^9) S:O(1)
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):
        def isValid(x,y):
            tmp=board[x][y]; board[x][y]='D'
            for i in range(9):
                if board[i][y]==tmp: return False
            for i in range(9):
                if board[x][i]==tmp: return False
            for i in range(3):
                for j in range(3):
                    if board[(x/3)*3+i][(y/3)*3+j]==tmp: return False
            board[x][y]=tmp
            return True
        def dfs(board):
            for i in range(9):
                for j in range(9):
                    if board[i][j]=='.':
                        for k in '123456789':
                            board[i][j]=k
                            if isValid(i,j) and dfs(board):
                                return True
                            board[i][j]='.'
                        return False
            return True
        dfs(board)

Run Time: 1440 ms