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:
11110
11010
11000
00000
Answer: 1
Example 2:
11000
11000
00100
00011
Answer: 3
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
本质是求矩阵中连续区域的个数， 可以用BFS, DFS, 或者 Union Find来解。
Java: BFS
class Coordinate {
int x, y;
public Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
}
public class Solution {
public int numIslands(boolean[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int n = grid.length;
int m = grid[0].length;
int islands = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j]) {
markByBFS(grid, i, j);
islands++;
}
}
}
return islands;
}
private void markByBFS(boolean[][] grid, int x, int y) {
int[] directionX = {0, 1, -1, 0};
int[] directionY = {1, 0, 0, -1};
Queue<Coordinate> queue = new LinkedList<>();
queue.offer(new Coordinate(x, y));
grid[x][y] = false;
while (!queue.isEmpty()) {
Coordinate coor = queue.poll();
for (int i = 0; i < 4; i++) {
Coordinate adj = new Coordinate(
coor.x + directionX[i],
coor.y + directionY[i]
);
if (!inBound(adj, grid)) {
continue;
}
if (grid[adj.x][adj.y]) {
grid[adj.x][adj.y] = false;
queue.offer(adj);
}
}
}
}
private boolean inBound(Coordinate coor, boolean[][] grid) {
int n = grid.length;
int m = grid[0].length;
return coor.x >= 0 && coor.x < n && coor.y >= 0 && coor.y < m;
}
}
Java: DFS
public class Solution {
private int m, n;
public void dfs(boolean[][] grid, int i, int j) {
if (i < 0 || i >= m || j < 0 || j >= n) return;
if (grid[i][j]) {
grid[i][j] = false;
dfs(grid, i - 1, j);
dfs(grid, i + 1, j);
dfs(grid, i, j - 1);
dfs(grid, i, j + 1);
}
}
public int numIslands(boolean[][] grid) {
m = grid.length;
if (m == 0) return 0;
n = grid[0].length;
if (n == 0) return 0;
int ans = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!grid[i][j]) continue;
ans++;
dfs(grid, i, j);
}
}
return ans;
}
}
Java: Union Find
class UnionFind {
private int[] father = null;
private int count;
private int find(int x) {
if (father[x] == x) {
return x;
}
return father[x] = find(father[x]);
}
public UnionFind(int n) {
// initialize your data structure here.
father = new int[n];
for (int i = 0; i < n; ++i) {
father[i] = i;
}
}
public void connect(int a, int b) {
int root_a = find(a);
int root_b = find(b);
if (root_a != root_b) {
father[root_a] = root_b;
count --;
}
}
public int query() {
return count;
}
public void set_count(int total) {
count = total;
}
}
public class Solution {
public int numIslands(boolean[][] grid) {
int count = 0;
int n = grid.length;
if (n == 0)
return 0;
int m = grid[0].length;
if (m == 0)
return 0;
UnionFind union_find = new UnionFind(n * m);
int total = 0;
for(int i = 0;i < grid.length; ++i)
for(int j = 0;j < grid[0].length; ++j)
if (grid[i][j])
total ++;
union_find.set_count(total);
for(int i = 0;i < grid.length; ++i)
for(int j = 0;j < grid[0].length; ++j)
if (grid[i][j]) {
if (i > 0 && grid[i - 1][j]) {
union_find.connect(i * m + j, (i - 1) * m + j);
}
if (i < n - 1 && grid[i + 1][j]) {
union_find.connect(i * m + j, (i + 1) * m + j);
}
if (j > 0 && grid[i][j - 1]) {
union_find.connect(i * m + j, i * m + j - 1);
}
if (j < m - 1 && grid[i][j + 1]) {
union_find.connect(i * m + j, i * m + j + 1);
}
}
return union_find.query();
}
}
Python: BFS
class Solution:
def numIslands(self, grid):
m = len(grid)
if m == 0:
return 0
n = len(grid[0])
visit = [[False for i in range(n)]for j in range(m)]
def check(x, y):
if x >= 0 and x<m and y>= 0 and y< n and grid[x][y] and visit[x][y] == False:
return True
def bfs(x,y):
nbrow = [1,0,-1,0]
nbcol = [0,1,0,-1]
q=[(x,y)]
while len(q) > 0:
x = q[0][0]
y = q[0][1]
q.pop(0)
for k in range(4):
newx = x + nbrow[k]
newy = y + nbcol[k]
if check(newx, newy):
visit[newx][newy] = True
q.append((newx,newy))
count = 0
for row in range(m):
for col in range(n):
if check(row,col):
visit[row][col] = True
bfs(row,col)
count+=1
return count
Python: DFS
class Solution:
def numIslands(self, grid):
if not grid:
return 0
row = len(grid)
col = len(grid[0])
count = 0
for i in xrange(row):
for j in xrange(col):
if grid[i][j] == ‘1‘:
self.dfs(grid, row, col, i, j)
count += 1
return count
def dfs(self, grid, row, col, x, y):
if grid[x][y] == ‘0‘:
return
grid[x][y] = ‘0‘
if x != 0:
self.dfs(grid, row, col, x - 1, y)
if x != row - 1:
self.dfs(grid, row, col, x + 1, y)
if y != 0:
self.dfs(grid, row, col, x, y - 1)
if y != col - 1:
self.dfs(grid, row, col, x, y + 1)
Python: DFS
class Solution:
def numIslands(self, grid):
m = len(grid)
if m == 0:
return 0
n = len(grid[0])
visit = [[False for i in range(n)]for j in range(m)]
def check(x, y):
if x >= 0 and x<m and y>= 0 and y< n and grid[x][y] and visit[x][y] == False:
return True
def dfs(x,y):
nbrow = [1,0,-1,0]
nbcol = [0,1,0,-1]
for k in range(4):
newx = x + nbrow[k]
newy = y + nbcol[k]
if check(newx, newy):
visit[newx][newy] = True
dfs(newx,newy)
count = 0
for row in range(m):
for col in range(n):
if check(row,col):
visit[row][col] = True
dfs(row,col)
count+=1
return count
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