力扣21 打卡17 设计相邻元素求和服务

思路：

该方案通过构建一个字典，将每个元素值映射到其在二维数组中的坐标位置，以便快速查找。adjacentSum 方法根据指定元素的坐标，计算其上下左右相邻元素之和；diagonalSum 方法则计算该元素的四个对角线相邻元素之和。每个方法通过判断相邻坐标是否在数组边界内，确保不越界访问。初始化字典的时间复杂度为 (O(n^2))，每次查询操作为 (O(1))，满足题目要求的效率。

from typing import List, Dict, Tuple

class NeighborSum:
    def __init__(self, grid: List[List[int]]):
        self.grid = grid
        self.n = len(grid)
        # Creating a dictionary to map each value to its (row, col) position in the grid
        self.value_to_position: Dict[int, Tuple[int, int]] = {}
        for i in range(self.n):
            for j in range(self.n):
                self.value_to_position[grid[i][j]] = (i, j)

    def adjacentSum(self, value: int) -> int:
        if value not in self.value_to_position:
            return 0
        row, col = self.value_to_position[value]
        adjacent_sum = 0
        # Checking all four possible adjacent cells (top, left, right, bottom)
        for dr, dc in [(-1, 0), (0, -1), (0, 1), (1, 0)]:
            new_row, new_col = row + dr, col + dc
            if 0 <= new_row < self.n and 0 <= new_col < self.n:
                adjacent_sum += self.grid[new_row][new_col]
        return adjacent_sum

    def diagonalSum(self, value: int) -> int:
        if value not in self.value_to_position:
            return 0
        row, col = self.value_to_position[value]
        diagonal_sum = 0
        # Checking all four diagonal cells (top-left, top-right, bottom-left, bottom-right)
        for dr, dc in [(-1, -1), (-1, 1), (1, -1), (1, 1)]:
            new_row, new_col = row + dr, col + dc
            if 0 <= new_row < self.n and 0 <= new_col < self.n:
                diagonal_sum += self.grid[new_row][new_col]
        return diagonal_sum