# 835. Image Overlap 图像重叠

@TOC

## # 题目描述

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 <= A.length = A[0].length = B.length = B[0].length <= 30
• 0 <= A[i][j], B[i][j] <= 1

## # 解题方法

Python代码如下：

``````class Solution:
def largestOverlap(self, A, B):
"""
:type A: List[List[int]]
:type B: List[List[int]]
:rtype: int
"""
N = len(A)
LA = [(xi, yi) for xi in range(N) for yi in range(N) if A[xi][yi]]
LB = [(xi, yi) for xi in range(N) for yi in range(N) if B[xi][yi]]
d = collections.Counter([(x1 - x2, y1 - y2) for (x1, y1) in LA for (x2, y2) in LB])
return max(d.values() or [0])
``````

``````class Solution {
public:
int largestOverlap(vector<vector<int>>& A, vector<vector<int>>& B) {
map<pair<int, int>, int> move;
vector<pair<int, int>> A1s, B1s;
for (int i = 0; i < A.size(); ++i) {
for (int j = 0; j < A[0].size(); ++j) {
if (A[i][j])
A1s.push_back({i, j});
}
}
for (int i = 0; i < B.size(); ++i) {
for (int j = 0; j < B[0].size(); ++j) {
if (B[i][j])
B1s.push_back({i, j});
}
}
int res = 0;
for (auto a : A1s) {
for (auto b : B1s) {
pair<int, int> dir = make_pair(b.first - a.first, b.second - a.second);
move[dir]++;
res = max(res, move[dir]);
}
}
return res;
}
};
``````

``````class Solution {
public:
int largestOverlap(vector<vector<int>>& A, vector<vector<int>>& B) {
const int N = A.size();
unordered_map<int, int> move;
vector<int> A1s, B1s;
for (int i = 0; i < N * N; ++i) {
if (A[i / N][i % N])
A1s.push_back((i / N) * 100 + i % N);
if (B[i / N][i % N])
B1s.push_back((i / N) * 100 + i % N);
}
int res = 0;
for (auto a : A1s) {
for (auto b : B1s) {
move[b - a]++;
res = max(res, move[b - a]);
}
}
return res;
}
};

``````

https://blog.csdn.net/zjucor/article/details/80298134 https://leetcode.com/problems/image-overlap/discuss/150504/Python-Easy-Logic https://leetcode.com/problems/image-overlap/discuss/130623/C%2B%2BJavaPython-Straight-Forward

## # 日期

2018 年 9 月 10 日 —— 教师节快乐！ 2018 年 12 月 28 日 —— 元旦假期到了