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xiaoyongyong發表於2024-11-25
1878. Get Biggest Three Rhombus Sums in a Grid

You are given an m x n integer matrix grid​​​.

A rhombus sum is the sum of the elements that form the border of a regular rhombus shape in grid​​​. The rhombus must have the shape of a square rotated 45 degrees with each of the corners centered in a grid cell. Below is an image of four valid rhombus shapes with the corresponding colored cells that should be included in each rhombus sum:

Note that the rhombus can have an area of 0, which is depicted by the purple rhombus in the bottom right corner.

Return the biggest three distinct rhombus sums in the grid in descending order. If there are less than three distinct values, return all of them.

Example 1:

Input: grid = [[3,4,5,1,3],[3,3,4,2,3],[20,30,200,40,10],[1,5,5,4,1],[4,3,2,2,5]]
Output: [228,216,211]
Explanation: The rhombus shapes for the three biggest distinct rhombus sums are depicted above.
- Blue: 20 + 3 + 200 + 5 = 228
- Red: 200 + 2 + 10 + 4 = 216
- Green: 5 + 200 + 4 + 2 = 211

Example 2:

Input: grid = [[1,2,3],[4,5,6],[7,8,9]]
Output: [20,9,8]
Explanation: The rhombus shapes for the three biggest distinct rhombus sums are depicted above.
- Blue: 4 + 2 + 6 + 8 = 20
- Red: 9 (area 0 rhombus in the bottom right corner)
- Green: 8 (area 0 rhombus in the bottom middle)

Example 3:

Input: grid = [[7,7,7]]
Output: [7]
Explanation: All three possible rhombus sums are the same, so return [7].

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 50
  • 1 <= grid[i][j] <= 105
class Solution {
    int m;
    int n;
    TreeSet<Integer> list = new TreeSet<>(); 
    public int[] getBiggestThree(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        // 計算半徑
        int len = Math.min(m, n) / 2;
        // for radius 0~len
        for(int radius = 0; radius <= len; radius++) {
            // center start sx = radius  sy = radius
            int sx = radius, sy = radius;
            // center end ex = m - radius, ey = n - radius
            int ex = m - radius, ey = n - radius;
            // i in sx~ex
            for(int i = sx; i <= ex; i++) {
                for(int j = sy; j <= ey; j++) {
                    int sum = calculate(grid, i, j, radius);
                    if(sum >= 0) list.add(sum);
                }
            }
        }
        int size = Math.min(list.size(), 3);
        int[] result = new int[size];
        for(int i = 0; i < size; i++) {
            result[i] = list.pollLast();
        }
        return result;
    }

    // 對以(x, y)為中心，radius為半徑的菱形求sum
    private int calculate(int[][] grid, int x, int y, int radius) {
        // 上下左右4個角
        int top = x - radius, bottom = x + radius, left = y - radius, right = y + radius;
        // 超出範圍校驗
        if(top < 0 || bottom >= m || left < 0 || right >= n) return -1;
        // radius為0的直接返回
        if(radius == 0) return grid[x][y];
        
        int sum = 0;
        //對於4條邊分別進行計算
        // left top
        for(int i = 0; i <= radius; i++) {
            sum += grid[x - i][left + i];
        }
        // right top 
        for(int i = 0; i <= radius; i++) {
            sum += grid[x - i][right - i];
        }
        // left bottom
        for(int i = 0; i <= radius; i++) {
            sum += grid[x + i][left + i];
        }
        // right top 
        for(int i = 0; i <= radius; i++) {
            sum += grid[x + i][right - i];
        }
        // 四個角多算了一次，要減去
        return sum - grid[x][left] - grid[x][right] - grid[top][y] - grid[bottom][y];
    }
}

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