LC 835. Image Overlap

标签：square   color   差值   values   shm   inpu   run   lap   坐标   

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

Runtime: 103 ms, faster than 42.68% of Java online submissions for Image Overlap.

注意，求两个点的向量差值，唯一表示要把坐标分开来，

这一题中，坐标小于30，所以La，Lb中add了i/N * 100 + i% N, i/N 和 i % N分别是横纵坐标。

最后的表示是一个4位数，xxxx。前两位是横坐标，后两位是纵坐标。

class Solution {

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

LC 835. Image Overlap

标签：square   color   差值   values   shm   inpu   run   lap   坐标   

原文地址：https://www.cnblogs.com/ethanhong/p/10256336.html