leetcode750 - Number Of Corner Rectangles - medium

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]]
Output: 0
Explanation: Rectangles must have four distinct corners.
Note:
1. The number of rows and columns of grid will each be in the range [1, 200].
2. Each grid[i][j] will be either 0 or 1.
3. The number of 1s in the grid will be at most 6000.

数学题。
1.Map<Integer, Map<Integer, Integer>>。O(~边长^3)。
遍历每一行，考虑在前面定好的情况下插入当前行会怎么增加长方形数。在这一行里二重循环遍历所有列的组合[l,r]，对lr的位置都是1的有效组合，如果之前记录过前面所有行[l,r]也为1的次数cnt，那么直接把cnt拿过来就是此时新增的正方形数了。同时记得根据这行更新维护cnt的MAP<MAP>。

2.组合计算。O(~边长^3)。
两行两行的组合[u, d]一起遍历。二重循环内部再用一个列指针j扫过去，看这两行里，同一列正好都是1的点对有几个，存到计数器n里。那么这两行所有的正方形数，也就是这n个双点对任选其二的组合: Cn2 = n*(n-1)/2了。遍历完所有行行的组合就有答案了。

实现1：

class Solution {
    public int countCornerRectangles(int[][] grid) {
        Map<Integer, Map<Integer, Integer>> map = new HashMap<>();

        int ans = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int l = 0; l < grid[0].length; l++) {
                if (grid[i][l] == 0) {
                    continue;
                }
                for (int r = l + 1; r < grid[0].length; r++) {
                    if (grid[i][r] == 0) {
                        continue;
                    }
                    if (map.containsKey(l) && map.get(l).containsKey(r)) {
                        ans += map.get(l).get(r);
                    }
                    map.putIfAbsent(l, new HashMap<Integer, Integer>());
                    map.get(l).put(r, map.get(l).getOrDefault(r, 0) + 1);
                }
            }
        }
        return ans;
    }
}

实现2：

class Solution {
    public int countCornerRectangles(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int ans = 0;
        for (int u = 0; u < grid.length; u++) {
            for (int d = u + 1; d < grid.length; d++) {
                int cnt = 0;
                for (int j = 0; j < grid[0].length; j++) {
                    if (grid[u][j] == 1 && grid[d][j] == 1) {
                        cnt++;
                    }
                }
                ans += cnt * (cnt - 1) / 2;
            }
        }
        return ans;
    }
}

小优化：

class Solution {
    public int countCornerRectangles(int[][] grid) {
        List<List<Integer>> dense = new ArrayList<>();
        Map<Integer, Map<Integer, Integer>> map = new HashMap<>();

        for (int i = 0; i < grid.length; i++) {
            dense.add(new ArrayList<Integer>());
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    dense.get(i).add(j);
                }
            }
        }

        int ans = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int cnt1 = 0; cnt1 < dense.get(i).size(); cnt1++) {
                int l = dense.get(i).get(cnt1);
                for (int cnt2 = cnt1 + 1; cnt2 < dense.get(i).size(); cnt2++) {
                    int r = dense.get(i).get(cnt2);
                    if (map.containsKey(l) && map.get(l).containsKey(r)) {
                        ans += map.get(l).get(r);
                    }
                    map.putIfAbsent(l, new HashMap<Integer, Integer>());
                    map.get(l).put(r, map.get(l).getOrDefault(r, 0) + 1);
                }
            }
        }
        return ans;
    }
}

原文地址：https://www.cnblogs.com/jasminemzy/p/9654923.html