# 694. Number of Distinct Islands 不同岛屿的数量

@TOC

## # 题目描述

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Count the number of distinct islands. An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.

Example 1:

``````11000
11000
00011
00011
Given the above grid map, return 1.
``````

Example 2:

``````11011
10000
00001
11011
Given the above grid map, return 3.
``````

Notice that:

``````11
1
``````

and

`````` 1
11
``````

are considered different island shapes, because we do not consider reflection / rotation.

Note: The length of each dimension in the given grid does not exceed 50.

## # 解题方法

### # DFS

C++中可以使用set对`vector<int>`去重，但是unordered_set是不可以的。

C++代码如下：

``````class Solution {
public:
int numDistinctIslands(vector<vector<int>>& grid) {
if (!grid.size() || !grid[0].size()) return 0;
M = grid.size();
N = grid[0].size();
set<vector<int>> s;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (grid[i][j] == 1) {
vector<int> path;
dfs(grid, path, i, j, i, j);
if (!path.empty()) {
s.insert(path);
}
}
}
}
return s.size();
}
void dfs(vector<vector<int>>& grid, vector<int>& path, int sr, int sc, int r, int c) {
if (r < 0 || r >= M || c < 0 || c >= N || grid[r][c] == 0) return;
path.push_back((r - sr) * N + c - sc);
grid[r][c] = 0;
dfs(grid, path, sr, sc, r - 1, c);
dfs(grid, path, sr, sc, r + 1, c);
dfs(grid, path, sr, sc, r, c - 1);
dfs(grid, path, sr, sc, r, c + 1);
}
private:
int M = 0;
int N = 0;
};
``````

## # 日期

2019 年 9 月 21 日 —— 莫生气，我若气病谁如意