other 1878

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;// 计算半径int len = Math.min(m, n) / 2;// for radius 0~lenfor(int radius = 0; radius <= len; radius++) {// center start sx = radius  sy = radiusint sx = radius, sy = radius;// center end ex = m - radius, ey = n - radiusint ex = m - radius, ey = n - radius;// i in sx~exfor(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;}// 对以(x, y)为中心，radius为半径的菱形求sumprivate int calculate(int[][] grid, int x, int y, int radius) {// 上下左右4个角int top = x - radius, bottom = x + radius, left = y - radius, right = y + radius;// 超出范围校验if(top < 0 || bottom >= m || left < 0 || right >= n) return -1;// radius为0的直接返回if(radius == 0) return grid[x][y];int sum = 0;//对于4条边分别进行计算// left topfor(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 bottomfor(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];}// 四个角多算了一次，要减去return sum - grid[x][left] - grid[x][right] - grid[top][y] - grid[bottom][y];}
}