750. Number Of Corner Rectangles

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本文介绍了一种高效算法，用于在二维数组中查找所有四个角为1的矩形数量。通过固定两行并逐列检查形成轴对齐矩形的方法，详细解析了算法原理与实现，最终给出代码示例。

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Description

Given a grid where each entry is only 0 or 1, find the number of corner rectangles.

A corner rectangle is 4 distinct 1s on the grid that form an axis-aligned rectangle. Note that only the corners need to have the value 1. Also, all four 1s used must be distinct.

Example 1:

Input: grid =
[[1, 0, 0, 1, 0],
[0, 0, 1, 0, 1],
[0, 0, 0, 1, 0],
[1, 0, 1, 0, 1]]
Output: 1
Explanation: There is only one corner rectangle, with corners grid[1][2], grid[1][4], grid[3][2], grid[3][4].

Example 2:

Input: grid =
[[1, 1, 1],
[1, 1, 1],
[1, 1, 1]]
Output: 9
Explanation: There are four 2x2 rectangles, four 2x3 and 3x2 rectangles, and one 3x3 rectangle.

Example 3:

Input: grid =
[[1, 1, 1, 1]]
Output: 0
Explanation: Rectangles must have four distinct corners.

Note:

The number of rows and columns of grid will each be in the range [1, 200].
Each grid[i][j] will be either 0 or 1.
The number of 1s in the grid will be at most 6000.

Problem URL

Solution

给一个二维数组，找到数组中所有四个角是1的矩形的数量。

To find an axis-aligned rectangle, my idea is to fix two rows (or two columns) first, then check column by column to find “1” on both rows. Say you find n pairs, then just pick any 2 of those to form an axis-aligned rectangle (calculating how many in total is just high school math, hint: combination).

Code

class Solution {
    public int countCornerRectangles(int[][] grid) {
        if (grid == null || grid.length <= 1){
            return 0;
        }
        int res = 0;
        for (int i = 0; i < grid.length - 1; i++){
            for (int j = i + 1; j < grid.length; j++){
                int count = 0;
                for (int k = 0; k < grid[0].length; k++){
                    if (grid[i][k] == 1 && grid[j][k] == 1){
                        count++;
                    }
                }
                res += count * (count - 1) / 2;
            }
        }
        return res;
    }
}