[Leetcode小记] 3242. 设计相邻元素求和服务

题目：

代码：

class NeighborSum {
    private int n;
    private int[][] grid;
    private int[][] adjacent_sum;
    private int[][] diagonal_sum;
    private int[][] index_map;
    public NeighborSum(int[][] grid) {
        this.grid=grid;
        n=grid.length;
        adjacent_sum=new int[n][n];
        diagonal_sum=new int[n][n];
        index_map=new int[n*n][2];
        //int count=0;//以为是按顺序排列的 实际顺序可能打乱
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                index_map[grid[i][j]][0]=i;
                index_map[grid[i][j]][1]=j;
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i==0&&j==0){//考虑左上角顶点
                    adjacent_sum[i][j]=grid[i+1][j]+grid[i][j+1];
                    diagonal_sum[i][j]=grid[i+1][j+1];
                }
                else if(i==n-1&&j==n-1){//考虑右下角顶点
                    adjacent_sum[i][j]=grid[i-1][j]+grid[i][j-1];
                    diagonal_sum[i][j]=grid[i-1][j-1];
                }
                else if(i==0&&j==n-1){//考虑右上角顶点
                    adjacent_sum[i][j]=grid[i+1][j]+grid[i][j-1];
                    diagonal_sum[i][j]=grid[i+1][j-1];
                }
                else if(i==n-1&&j==0){//考虑左下角顶点
                    adjacent_sum[i][j]=grid[i-1][j]+grid[i][j+1];
                    diagonal_sum[i][j]=grid[i-1][j+1];
                }
                else if(i==0){//第一行除了左上右上顶点的数
                    adjacent_sum[i][j]=grid[i+1][j]+grid[i][j+1]+grid[i][j-1];
                    diagonal_sum[i][j]=grid[i+1][j+1]+grid[i+1][j-1];
                }
                else if(i==n-1){//最后一行除了左下右下顶点的数
                    adjacent_sum[i][j]=grid[i-1][j]+grid[i][j+1]+grid[i][j-1];
                    diagonal_sum[i][j]=grid[i-1][j-1]+grid[i-1][j+1];
                }
                else if(j==0){//第一列除了左上左下顶点的数
                    adjacent_sum[i][j]=grid[i-1][j]+grid[i+1][j]+grid[i][j+1];
                    diagonal_sum[i][j]=grid[i+1][j+1]+grid[i-1][j+1];
                }
                else if(j==n-1){//最后一列除了右上右下顶点的数
                    adjacent_sum[i][j]=grid[i-1][j]+grid[i+1][j]+grid[i][j-1];
                    diagonal_sum[i][j]=grid[i+1][j-1]+grid[i-1][j-1];
                }
                else{//非顶点、非第一行、非第一列、非最后一行、非最后一列的数
                    adjacent_sum[i][j]=grid[i-1][j]+grid[i+1][j]+grid[i][j-1]+grid[i][j+1];
                    diagonal_sum[i][j]=grid[i-1][j-1]+grid[i-1][j+1]+grid[i+1][j-1]+grid[i+1][j+1];
                }
            }
        }
    }
    
    public int adjacentSum(int value) {
        return adjacent_sum[index_map[value][0]][index_map[value][1]];
    }
    
    public int diagonalSum(int value) {
       return diagonal_sum[index_map[value][0]][index_map[value][1]];
    }
}

/**
 * Your NeighborSum object will be instantiated and called as such:
 * NeighborSum obj = new NeighborSum(grid);
 * int param_1 = obj.adjacentSum(value);
 * int param_2 = obj.diagonalSum(value);
 */

总结：

方法为If-Else大法，算法上不是最优，优化该代码可采用预处理。

时间复杂度：初始化 O(n^2)，其余 O(1) (n为grid的行数和列数)

空间复杂度：初始化 O(n^2)，其余 O(1)。