LWC 63: 750. Number Of Corner Rectangles

LWC 63: 750. Number Of Corner Rectangles

传送门：750. Number Of Corner Rectangles

Problem:

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 grid1[2], grid1[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.

思路：
暴力搜索，实际上如果矩形的纵向边被确定了，只要有两条以上自然能构成矩形，所以只需要遍历不同的两行，找能够构成纵向边的个数，再组合一波即可。

Java版本：

    public int countCornerRectangles(int[][] grid) {        int n = grid.length;        int m = grid[0].length;        int ret = 0;        for (int i = 0; i < n; ++i) {            for (int j = i + 1; j < n; ++j) {                int np = 0;                for (int k = 0; k < m; ++k) {                    if (grid[i][k] == 1 && grid[j][k] == 1) {                        np ++;                    }                }                ret += np * (np - 1) / 2;            }        }        return ret;    }

Python版本：

    def countCornerRectangles(self, grid):        """        :type grid: List[List[int]]        :rtype: int        """        n = len(grid)        m = len(grid[0])        res = 0        for i in xrange(n):            for j in xrange(i + 1, n):                np = 0                for k in xrange(m):                    if grid[i][k] and grid[j][k]:                        np += 1                res += np * (np - 1) / 2        return res