Re: [閒聊] 每日LeetCode已回收

看板Marginalman作者Rushia (みけねこ的鼻屎)時間3年前 (2022/10/28 00:34)推噓2(2推 0噓 2→)

留言4則, 3人參與討論串65/719 (看更多)

835. Image Overlap 給予兩個矩陣A和B，我們可以任意上下左右的位移該矩陣，判斷經過任意次位移後可得的最大重疊數（A[y][x] == B[y][x] == 1 我們說他重疊）。 https://assets.leetcode.com/uploads/2020/09/09/overlap1.jpg

Input: img1 = [[1,1,0],[0,1,0],[0,1,0]], img2 = [[0,0,0],[0,1,1],[0,0,1]] Output: 3 Explanation: We translate img1 to right by 1 unit and down by 1 unit. https://assets.leetcode.com/uploads/2020/09/09/overlap_step1.jpg

The number of positions that have a 1 in both images is 3 (shown in red). https://assets.leetcode.com/uploads/2020/09/09/overlap_step2.jpg

思路： 1.因為我想睡覺所以我們應該對兩個矩陣使用暴力法 2.用-n+1~n-1表示左到右、上到下的位移，然後直接比較A和B位移過後是否都是1 3.統計完的sum取大的值 JavaCode： class Solution { public int largestOverlap(int[][] A, int[][] B) { int n = A.length; int res = 0; for(int i = -n + 1; i < n; i++) for(int j = -n + 1; j < n; j++) { int sum = 0; for(int x = 0; x < n; x++) for(int y = 0; y < n; y++) if(x + i >= 0 && x + i < n && y + j >= 0 && y + j < n) sum += A[x + i][y + j] & B[x][y]; res = Math.max(res, sum); } return res; } } 晚安 https://i.imgur.com/nYPrAZB.png