• LC 835. Image Overlap


    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;
      }
    }
  • 相关阅读:
    Java实现多线程的四种实现方式
    电梯调度算法[转]
    带黑洞的随机游走问题
    深度学习印象
    使用jupyterthemes插件定制jupyter notebook界面
    tf.gfile
    中国象棋残局库构建[抄]
    Android(Linux)线路规程的使用
    Remote Displayer for Android V1.2
    Android开发资源汇总
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10256336.html
Copyright © 2020-2023  润新知