# 694. Number of Distinct Islands

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.

``````class Solution {
public:
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
int numDistinctIslands(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
unordered_set<string> res;
vector<vector<bool>> visited(m, vector<bool>(n, false));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1 && !visited[i][j]) {
set<string> s;
helper(grid, i, j, i, j, visited, s);
string t = "";
for (auto str : s) t += str + "_";
res.insert(t);
}
}
}
return res.size();
}
void helper(vector<vector<int>>& grid, int x0, int y0, int i, int j, vector<vector<bool>>& visited, set<string>& s) {
int m = grid.size(), n = grid[0].size();
visited[i][j] = true;
for (auto dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] == 0 || visited[x][y]) continue;
string str = to_string(x - x0) + "_" + to_string(y - y0);
s.insert(str);
helper(grid, x0, y0, x, y, visited, s);
}
}
};
``````

``````class Solution {
public:
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
int numDistinctIslands(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
set<vector<pair<int, int>>> res;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] != 1) continue;
vector<pair<int, int>> v;
helper(grid, i, j, i, j, v);
res.insert(v);
}
}
return res.size();
}
void helper(vector<vector<int>>& grid, int x0, int y0, int i, int j, vector<pair<int, int>>& v) {
int m = grid.size(), n = grid[0].size();
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] <= 0) return;
grid[i][j] *= -1;
v.push_back({i - x0, j - y0});
for (auto dir : dirs) {
helper(grid, x0, y0, i + dir[0], j + dir[1], v);
}
}
};
``````

``````class Solution {
public:
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
int numDistinctIslands(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
set<vector<pair<int, int>>> res;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] != 1) continue;
vector<pair<int, int>> v;
queue<pair<int, int>> q{{{i, j}}};
grid[i][j] *= -1;
while (!q.empty()) {
auto t = q.front(); q.pop();
for (auto dir : dirs) {
int x = t.first + dir[0], y = t.second + dir[1];
if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] <= 0) continue;
q.push({x, y});
grid[x][y] *= -1;
v.push_back({x - i, y - j});
}
}
res.insert(v);
}
}
return res.size();
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/694

Number of Islands

Number of Distinct Islands II

https://leetcode.com/problems/number-of-distinct-islands/

https://leetcode.com/problems/number-of-distinct-islands/discuss/150037/DFS-with-Explanations

https://leetcode.com/problems/number-of-distinct-islands/discuss/108474/JavaC%2B%2B-Clean-Code

https://leetcode.com/problems/number-of-distinct-islands/discuss/108475/Java-very-Elegant-and-concise-DFS-Solution(Beats-100)

LeetCode All in One 题目讲解汇总(持续更新中…)

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