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➤微信公众号:山青咏芝(shanqingyongzhi)
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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.)git
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.github
(Note also that a translation does not include any kind of rotation.)微信
What is the largest possible overlap?app
Example 1:ide
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: this
1 <= A.length = A[0].length = B.length = B[0].length <= 30
0 <= A[i][j], B[i][j] <= 1
给出两个图像 A
和 B
,A
和 B
为大小相同的二维正方形矩阵。(而且为二进制矩阵,只包含0和1)。spa
咱们转换其中一个图像,向左,右,上,或下滑动任何数量的单位,并把它放在另外一个图像的上面。以后,该转换的重叠是指两个图像都具备 1 的位置的数目。code
(请注意,转换不包括向任何方向旋转。)htm
最大可能的重叠是什么?
示例 1:
输入:A = [[1,1,0], [0,1,0], [0,1,0]] B = [[0,0,0], [0,1,1], [0,0,1]] 输出:3 解释: 将 A 向右移动一个单位,而后向下移动一个单位。
注意:
1 <= A.length = A[0].length = B.length = B[0].length <= 30
0 <= A[i][j], B[i][j] <= 1
288ms
1 class Solution { 2 func largestOverlap(_ A: [[Int]], _ B: [[Int]]) -> Int { 3 var LA = [Int](), LB = [Int]() 4 let N = A.count 5 var count = [Int: Int]() 6 for i in 0..<N*N { 7 if A[i/N][i%N] == 1 { 8 LA.append( i/N * 100 + i % N) 9 } 10 if B[i/N][i%N] == 1 { 11 LB.append( i/N * 100 + i % N) 12 } 13 } 14 15 for i in LA { 16 for j in LB { 17 count[i-j, default: 0] += 1 18 } 19 } 20 21 var res = 0 22 for (k, v) in count { 23 res = max(res, v) 24 } 25 return res 26 } 27 }
1 class Solution { 2 func largestOverlap(_ A: [[Int]], _ B: [[Int]]) -> Int { 3 var res:Int = 0 4 var n:Int = A.count 5 var listA:[Int] = [Int]() 6 var listB:[Int] = [Int]() 7 var diffCnt:[Int:Int] = [Int:Int]() 8 for i in 0..<(n * n) 9 { 10 if A[i / n][i % n] == 1 11 { 12 listA.append(i / n * 100 + i % n) 13 } 14 if B[i / n][i % n] == 1 15 { 16 listB.append(i / n * 100 + i % n) 17 } 18 } 19 for a in listA 20 { 21 for b in listB 22 { 23 diffCnt[a - b,default:0] += 1 24 } 25 } 26 for diff in diffCnt 27 { 28 res = max(res, diff.value) 29 } 30 return res 31 } 32 }