835. Image Overlap

开源项目
1

Two images A and B are given, represented as binary, square matrices of the same size. (A binary matrix has only 0s and 1s as values.)

We translate one image however we choose (sliding it left, right, up, or down any number of units), and place it on top of the other image. After, the overlap of this translation is the number of positions that have a 1 in both images.

(Note also that a translation does not include any kind of rotation.)

What is the largest possible overlap?

Example 1:

Input: A = [[1,1,0], [0,1,0], [0,1,0]] B = [[0,0,0], [0,1,1], [0,0,1]]

Output: 3

Explanation: We slide A to right by 1 unit and down by 1 unit.

Notes:

  1. 1 <= A.length = A[0].length = B.length = B[0].length <= 30
  2. 0 <= A[i][j], B[i][j] <= 1

思路

只要A和B中1的相对位置相同，那么平移后就可以覆盖，当值等于1时，把位置存起来成LA，LB，然后对LA和LB中的位置进行想减，把减后的值进行叠加，最后挑选出减后相同值最多的情况。

A中每个元素的位置都是不同的，当存入LA时也有一个计算方法让LA的每个值都与众不同：i / N * 100 + i % N，我选用100是因为数组的最大长度为30，只要大于2*N - 1都可以

class Solution {
      public int largestOverlap(int[][] A, int[][] B) {
        int N = A.length;
        List<Integer> LA = new ArrayList<>(),  LB = new ArrayList<>();
        HashMap<Integer, Integer> count = new HashMap<>();
        for (int i = 0; i < N * N; ++i)
            if (A[i / N][i % N] == 1)
                LA.add(i / N * 100 + i % N);
        for (int i = 0; i < N * N; ++i)
            if (B[i / N][i % N] == 1)
                LB.add(i / N * 100 + i % N);
        for (int i : LA) for (int j : LB)
                count.put(i - j, count.getOrDefault(i - j, 0) + 1);
        int res = 0;
        for (int i : count.values())
            res = Math.max(res, i);
        return res;
    }
}