Leetcode 840. Magic Squares In Grid

Posted 2021-07-04 SnailTyan

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

**解析：**根据三阶幻方的定义，一项项检查即可，满足条件+1。Version 1是针对3*3的，Version 2进行了一些修改，减少了代码量，Version 3是针对k*k的通用版本。

• Version 1

class Solution:
    def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        columns = len(grid[0])
        count = 0
        for i in range(rows - 2):
            for j in range(columns - 2):
                if self.check(grid, i, j):
                    count += 1
        return count


    def check(self, grid, x, y):
        temp = set(grid[x][y:y+3] + grid[x+1][y:y+3] + grid[x+2][y:y+3])
        if len(temp) != 9:
            return False
        for i in range(x, x+3):
            for j in range(y, y+3):
                if grid[i][j] == 0 or grid[i][j] > 9:
                    return False
        x1 = sum(grid[x][y:y+3])
        x2 = sum(grid[x+1][y:y+3])
        if x2 != x1:
            return False
        x3 = sum(grid[x+2][y:y+3])
        if x3 != x1:
            return False
        x4 = grid[x][y] + grid[x+1][y] + grid[x+2][y]
        if x4 != x1:
            return False
        x5 = grid[x][y+1] + grid[x+1][y+1] + grid[x+2][y+1]
        if x5 != x1:
            return False
        x6 = grid[x][y+2] + grid[x+1][y+2] + grid[x+2][y+2]
        if x6 != x1:
            return False
        x7 = grid[x][y] + grid[x+1][y+1] + grid[x+2][y+2]
        if x7 != x1:
            return False
        x8 = grid[x][y+2] + grid[x+1][y+1] + grid[x+2][y]
        if x8 != x1:
            return False
        return True

• Version 2

class Solution:
    def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        columns = len(grid[0])
        count = 0
        for i in range(rows - 2):
            for j in range(columns - 2):
                if self.check(grid, i, j):
                    count += 1
        return count


    def check(self, grid, x, y):
        temp = {i: 0 for i in range(1, 10)}
        for i in range(x, x+3):
            for j in range(y, y+3):
                if grid[i][j] not in temp or temp[grid[i][j]] == 1:
                    return False
                temp[grid[i][j]] = 1

        temp = grid[x][y] + grid[x+1][y+1] + grid[x+2][y+2]
        if temp != grid[x][y+2] + grid[x+1][y+1] + grid[x+2][y]:
            return False
        for i in range(3):
            if temp != sum(grid[x+i][y:y+3]):
                return False
            if temp != grid[x][y+i] + grid[x+1][y+i] + grid[x+2][y+i]:
                return False
        return True

• Version 3

class Solution:
    def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        columns = len(grid[0])
        count = 0
        k = 3
        for i in range(rows-k+1):
            for j in range(columns-k+1):
                if self.check(grid, i, j, k):
                    count += 1
        return count


    def check(self, grid, x, y, k):
        temp = {i: 0 for i in range(1, k*k+1)}
        for i in range(x, x+k):
            for j in range(y, y+k):
                if grid[i][j] not in temp or temp[grid[i][j]] == 1:
                    return False
                temp[grid[i][j]] = 1

        main_diagonal = 0
        secondary_diagonal = 0
        for i in range(k):
            main_diagonal += grid[x+i][y+i]
            secondary_diagonal += grid[x+i][y+k-i-1]
        if main_diagonal != secondary_diagonal:
            return False
        for i in range(k):
            if main_diagonal != sum(grid[x+i][y:y+k]):
                return False
            if main_diagonal != sum([grid[x+j][y+i] for j in range(k)]):
                return False
        return True

Reference

1. https://leetcode.com/problems/magic-squares-in-grid/