【leetcode】1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

题目如下：

Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbours of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighboors if they share one edge.

Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.

Binary matrix is a matrix with all cells equal to 0 or 1 only.

Zero matrix is a matrix with all cells equal to 0.

Example 1:

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

Example 2:

Input: mat = [[0]]
Output: 0
Explanation: Given matrix is a zero matrix. We don‘t need to change it.

Example 3:

Input: mat = [[1,1,1],[1,0,1],[0,0,0]]
Output: 6

Example 4:

Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix can‘t be a zero matrix

Constraints:

• m == mat.length
• n == mat[0].length
• 1 <= m <= 3
• 1 <= n <= 3
• mat[i][j] is 0 or 1.

解题思路：最大就是3*3的矩阵，BFS把每种情况都试一遍就好了。

代码如下：

class Solution(object):
    def minFlips(self, mat):
        """
        :type mat: List[List[int]]
        :rtype: int
        """
        import copy
        def isAllZero(grid):
            count = 0
            for i in grid:
                count += sum(i)
            return count == 0

        def encode(grid):
            grid_s = ‘‘
            for i in range(len(grid)):
                for j in range(len(grid[i])):
                    grid_s += str(grid[i][j])
                if i != len(grid) - 1 : grid_s += ‘#‘
            return grid_s
        def decode(grid_s):
            gl = grid_s.split(‘#‘)
            grid = []
            for i in gl:
                tl = []
                for j in list(i):
                    tl.append(int(j))
                grid.append(tl)
            return grid

        res = float(‘inf‘)
        queue = [(encode(mat),0)]
        dic = {}
        dic[encode(mat)] = 0

        while len(queue) > 0:
            gs,step = queue.pop(0)
            grid = decode(gs)
            if isAllZero(grid):
                res = min(res,step)
                continue
            elif res <= step:
                continue
            for i in range(len(grid)):
                for j in range(len(grid[i])):
                    #if grid[i][j] == 0: continue
                    new_grid = copy.deepcopy(grid)
                    if new_grid[i][j] == 1:
                        new_grid[i][j] = 0
                    else:new_grid[i][j] = 1
                    directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
                    for (x, y) in directions:
                        if x + i >= 0 and x + i < len(grid) and y + j >= 0 and y + j < len(grid[0]):
                            if new_grid[x + i][y + j] == 0:
                                new_grid[x + i][y + j] = 1
                            else:
                                new_grid[x + i][y + j] = 0
                    encode_Str = encode(new_grid)
                    if encode_Str not in dic or dic[encode_Str] > step + 1:
                        queue.append((encode(new_grid), step + 1))
                        dic[encode_Str] = step + 1
        return res if res != float(‘inf‘) else -1

原文地址：https://www.cnblogs.com/seyjs/p/12026296.html