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• Leetcode 3548. Equal Sum Grid Partition II
  • 1. 解题思路
  • 2. 代码实现

• 题目链接：3548. Equal Sum Grid Partition II

1. 解题思路

这一题是题目3546. Equal Sum Grid Partition I的进阶版本，不过本质上还是差不多的。

相较于题目3546，这里的改动是可以允许至多一个元素的清零，但不能使得区域不连续。

因此，我们就是分别要在横向和纵向考察以每一个位置进行切分时，两侧的元素差，然后考察较多的那一部分是否恰好有与差值完全相同的元素，且这个元素如果被去除的话是否会将对应的区域完全切断开来，这个我们只需要分类考察一下即可。

2. 代码实现

给出python代码实现如下：

class Solution:def canPartitionGrid(self, grid: List[List[int]]) -> bool:n, m = len(grid), len(grid[0])cnt = defaultdict(int)tot = 0for i in range(n):for j in range(m):cnt[grid[i][j]] += 1tot += grid[i][j]_cnt = defaultdict(int)_tot = 0for i in range(n-1):for j in range(m):_cnt[grid[i][j]] += 1_tot += grid[i][j]delta = 2 * _tot - totif delta == 0:return Trueelif delta > 0:if _cnt[delta] > 0 and i > 0 and m > 1:return Trueelif i > 0 and m == 1 and (grid[0][0] == delta or grid[i][0] == delta):return Trueelif i == 0 and (grid[0][0] == delta or grid[0][-1] == delta):return Trueelse:if cnt[-delta] - _cnt[-delta] > 0 and i < n-2 and m > 1:return Trueelif i < n-2 and m == 1 and (grid[-1][0] == -delta or grid[i+1][0] == -delta):return Trueelif i == n-2 and (grid[-1][0] == -delta or grid[-1][-1] == -delta):return True_cnt = defaultdict(int)_tot = 0for j in range(m-1):for i in range(n):_cnt[grid[i][j]] += 1_tot += grid[i][j]delta = 2 * _tot - totif delta == 0:return Trueelif delta > 0:if _cnt[delta] > 0 and j > 0 and n > 1:return Trueelif j > 0 and n == 1 and (grid[0][0] == delta or grid[0][j] == delta):return Trueelif j == 0 and (grid[0][0] == delta or grid[-1][0] == delta):return Trueelse:if cnt[-delta] - _cnt[-delta] > 0 and j < m-2 and n > 1:return Trueelif j < m-2 and n == 1 and (grid[0][-1] == -delta or grid[0][j+1] == -delta):return Trueelif j == m-2 and (grid[0][-1] == -delta or grid[-1][-1] == -delta):return Truereturn False      

提交代码评测得到：耗时847ms，占用内存57.9MB。