LC 835. Image Overlap
Two images? 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,? [0,0]] ? B = [[0,1],1]] Output: 3 Explanation: We slide A to right by 1 unit and down by 1 unit. Notes:?
? 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; } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |