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  

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