leetcode-166周赛-5282-转化为全0矩阵的最小反转次数

题目描述：

方法一：暴力BFS

class Solution:
    def minFlips(self, mat) -> int:
        R, C = len(mat), len(mat[0])

        def helper(mat):
            return tuple(tuple(row) for row in mat)

        visited = set()
        queue = collections.deque()
        mat_ = helper(mat)
        visited.add(mat_)
        queue.append(mat)
        ans = 0
        while queue:
            for _ in range(len(queue)):
                cur = queue.popleft()
                if sum(sum(row) for row in cur) == 0:
                    return ans
                for i in range(R):
                    for j in range(C):
                        cur1 = [row[:] for row in cur] #列表的复制一定要深copy
                        for i_, j_ in [[i, j], [i - 1, j], [i + 1, j], [i, j + 1], [i, j - 1]]:
                            if 0 <= i_ < R and 0 <= j_ < C:
                                cur1[i_][j_] ^= 1
                        cur_ = helper(cur1)
                        if cur_ not in visited:
                            visited.add(cur_)
                            queue.append(cur1)
            ans += 1
        return -1

方法二：暴力枚举

class Solution:
    def minFlips(self, mat) -> int:
        R,C = len(mat),len(mat[0])
        N = R*C
        ans = R*C + 1
        for i in itertools.product(range(2),repeat = N):
            bns = sum(i)
            if bns >= ans:
                continue
            r = c = 0
            B = [row[:] for row in mat]
            for adj in i:
                if adj:
                    for i_,j_ in [[r,c],[r+1,c],[r-1,c],[r,c+1],[r,c-1]]:
                        if 0 <= i_ < R and 0 <= j_ < C:
                            B[i_][j_] ^= 1
                c += 1
                if c == C:
                    c = 0
                    r += 1
            if sum(sum(row)for row in B) == 0:
                ans = bns
        return ans if ans < N+1 else -1