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：
代码语言：javascript复制 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：
代码语言：javascript复制 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 
