LeetCode-37-Sudoku Solver， list转字符串join，回溯_java 分隔字符 leetcode join-优快云博客

本文链接：https://blog.youkuaiyun.com/qdbszsj/article/details/77972947

本文介绍了一种使用深度优先搜索(DFS)策略解决数独问题的算法实现。该算法通过递归地尝试填入数字并检查是否违反数独规则来寻找解决方案。文章详细解释了如何判断填充的数字是否合法，并提供了完整的Python代码示例。

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class Solution(object):
    count=0
    ans=[[0 for x in range(9)]for y in range(9)]
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        B=[[0 for x in range(9)]for y in range(9)]
        for i in range(9):
            for j in range(9):
                if board[i][j]=='.':continue
                B[i][j]=ord(board[i][j])-ord('0')
        for i in range(9):
            for j in range(9):
                if B[i][j] == 0:
                    flag=self.dfs(B, i, j)
                    if flag:
                        #board=B
                        self.copyBoard(board,B)
                        #print B
                        #print board
                        return
                    break

    def copyBoard(self,board,B):
        for i in range(9):
            for j in range(9):
                B[i][j]=chr(ord("0")+B[i][j])
        for i in range(9):
            board[i]="".join(B[i])



    def dfs(self, B, x, y):
        for num in range(1, 10):
            B[x][y] = num
            if self.judgePoint(B, x, y) == False:
                B[x][y] = 0
                continue
            nextx = x
            nexty = y
            while nextx < 9:
                nexty = nexty + 1
                if nexty == 9:
                    nexty = 0
                    nextx += 1
                    if nextx == 9: return True
                if B[nextx][nexty] == 0:
                    break
            flag=self.dfs(B, nextx, nexty)
            if flag:
                return True
            B[x][y] = 0
        return False

    def judgePoint(self, board, x, y):
        area = 0
        for i in range(9):
            if i!=y and board[x][i]==board[x][y]:return False
            if i!=x and board[i][y]==board[x][y]:return False
            px = x / 3 * 3 + i / 3
            py = y / 3 * 3 + i - i / 3 * 3
            c = board[px][py]
            if c == 0: continue
            num = c
            if area & (1 << num) != 0:
                return False
            area |= 1 << num
        return True