原题链接在这里：https://leetcode.com/problems/number-of-corner-rectangles/
题目：
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
gridwill each be in the range[1, 200]. - Each
grid[i][j]will be either0or1. - The number of
1s in the grid will be at most6000.
题解：
When there is a new row, within this row, row[c1] and row[c2] are both 1.
Then number of newly constructed rectanges are number of previous rows having both exact same two columns mark as 1.
To maintain number of two 1 on c1 and c2, use a hash function c1*200+c2.
Time Complexity: O(r*c^2). r = grid.length. c = grid[0].length.
Space: O(c^2).
AC Java:
1 class Solution {
2 public int countCornerRectangles(int[][] grid) {
3 int res = 0;
4 HashMap<Integer, Integer> hm = new HashMap<>();
5 for(int [] row : grid){
6 for(int c1 = 0; c1<row.length; c1++){
7 if(row[c1] == 1){
8 for(int c2 = c1+1; c2<row.length; c2++){
9 if(row[c2] == 1){
10 int hash = c1*200+c2;
11 int count = hm.getOrDefault(hash, 0);
12 res += count;
13 hm.put(hash, count+1);
14 }
15 }
16 }
17 }
18 }
19
20 return res;
21 }
22 }
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11438546.html