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.
Discuss
暴力枚举+轻微减枝
class Solution {
public:
int search(vector<vector<int>>& grid, int x, int y) {
int n = grid.size();
int m = grid[0].size();
int cnt = 0;
for (int i = 1; i < n - x; ++i) {
if(grid[x+i][y] == 0) continue;
for (int j = 1; j < m - y; ++j) {
int x1, y1;
x1 = x;
y1 = y + j;
if (grid[x1][y1] == 0) continue;
x1 = x + i;
y1 = y + j;
if (grid[x1][y1] == 0) continue;
x1 = x + i;
y1 = y;
if (grid[x1][y1] == 0) continue;
cnt++;
}
}
return cnt;
}
int countCornerRectangles(vector<vector<int>>& grid) {
int n = grid.size();
int m = grid[0].size();
if (n == 1) return 0;
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
ans += search(grid, i, j);
}
}
}
return ans;
}
};