Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input: 11110 11010 11000 00000 Output: 1
Example 2:
Input: 11000 11000 00100 00011 Output: 3
Related Topics:
Depth-first Search, Breadth-first Search, Union Find
Similar Questions:
- Surrounded Regions (Medium)
- Walls and Gates (Medium)
- Number of Islands II (Hard)
- Number of Connected Components in an Undirected Graph (Medium)
- Number of Distinct Islands (Medium)
- Max Area of Island (Medium)
// OJ: https://leetcode.com/problems/number-of-islands/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class UnionFind {
vector<int> id;
public:
UnionFind(int N) : id(N) {
iota(begin(id), end(id), 0);
}
void connect(int x, int y) {
int a = find(x), b = find(y);
if (a == b) return;
id[a] = b;
}
int find(int x) {
return id[x] == x ? x : (id[x] = find(id[x]));
}
};
class Solution {
int M, N;
int key(int x, int y) { return x * N + y; }
int dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
public:
int numIslands(vector<vector<char>>& A) {
if (A.empty() || A[0].empty()) return 0;
M = A.size(), N = A[0].size();
UnionFind uf(M * N);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == '0') continue;
for (auto &dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || y < 0 || x >= M || y >= N || A[x][y] == '0') continue;
uf.connect(key(i, j), key(x, y));
}
}
}
unordered_set<int> s;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == '0') continue;
s.insert(uf.find(key(i, j)));
}
}
return s.size();
}
};// OJ: https://leetcode.com/problems/number-of-islands/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
int M, N;
int dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
void dfs(vector<vector<char>>& A, int x, int y) {
if (x < 0 || y < 0 || x >= M || y >= N || A[x][y] != '1') return;
A[x][y] = 'x';
for (auto &dir : dirs) dfs(A, x + dir[0], y + dir[1]);
}
public:
int numIslands(vector<vector<char>>& A) {
if (A.empty() || A[0].empty()) return 0;
int ans = 0;
M = A.size(), N = A[0].size();
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] != '1') continue;
++ans;
dfs(A, i, j);
}
}
return ans;
}
};// OJ: https://leetcode.com/problems/number-of-islands/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
int M, N;
int dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
void bfs(vector<vector<char>> &A, int x, int y) {
queue<pair<int, int>> q;
q.emplace(x, y);
A[x][y] = 'x';
while (q.size()) {
x = q.front().first, y = q.front().second;
q.pop();
for (auto &dir : dirs) {
int i = x + dir[0], j = y + dir[1];
if (i < 0 || j < 0 || i >= M || j >= N || A[i][j] != '1') continue;
A[i][j] = 'x';
q.emplace(i, j);
}
}
}
public:
int numIslands(vector<vector<char>>& A) {
if (A.empty() || A[0].empty()) return 0;
int ans = 0;
M = A.size(), N = A[0].size();
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] != '1') continue;
++ans;
bfs(A, i, j);
}
}
return ans;
}
};