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.
Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
Example 1:
[[0,0,1,0,0,0,0,1,0,0,0,0,0], [0,0,0,0,0,0,0,1,1,1,0,0,0], [0,1,1,0,1,0,0,0,0,0,0,0,0], [0,1,0,0,1,1,0,0,1,0,1,0,0], [0,1,0,0,1,1,0,0,1,1,1,0,0], [0,0,0,0,0,0,0,0,0,0,1,0,0], [0,0,0,0,0,0,0,1,1,1,0,0,0], [0,0,0,0,0,0,0,1,1,0,0,0,0]]
Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.
Example 2:
[[0,0,0,0,0,0,0,0]]
Given the above grid, return 0.
Note: The length of each dimension in the given grid does not exceed 50.
DFS:
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
def dfs(i, j):
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 0:
return 0
grid[i][j] = 0
res = 1
for x, y in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
res += dfs(i + x, j + y)
return res
m, n = len(grid), len(grid[0])
res = 0
for i in range(m):
for j in range(n):
t = dfs(i, j)
res = max(res, t)
return resUnion find:
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
p = list(range(m * n))
size = [1] * (m * n)
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
for i in range(m):
for j in range(n):
if grid[i][j] == 1:
if i < m - 1 and grid[i + 1][j] == 1:
a, b = find(i * n + j), find((i + 1) * n + j)
if a != b:
size[a] += size[b]
p[b] = a
if j < n - 1 and grid[i][j + 1] == 1:
a, b = find(i * n + j), find(i * n + j + 1)
if a != b:
size[a] += size[b]
p[b] = a
res = 0
for i in range(m):
for j in range(n):
if grid[i][j] == 1:
res = max(res, size[i * n + j])
return resDFS:
class Solution {
private int[][] directions = {{0, 1}, {0, - 1}, {1, 0}, {-1, 0}};
public int maxAreaOfIsland(int[][] grid) {
int m = grid.length, n = grid[0].length;
int res = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int t = dfs(grid, i, j, m, n);
res = Math.max(res, t);
}
}
return res;
}
private int dfs(int[][] grid, int i, int j, int m, int n) {
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
return 0;
}
grid[i][j] = 0;
int res = 1;
for (int[] direction : directions) {
res += dfs(grid, i + direction[0], j + direction[1], m, n);
}
return res;
}
}Union find:
class Solution {
private int[] p;
private int[] size;
public int maxAreaOfIsland(int[][] grid) {
int m = grid.length, n = grid[0].length;
p = new int[m * n];
size = new int[m * n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
p[i * n + j] = i * n + j;
size[i * n + j] = 1;
}
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
if (i < m - 1 && grid[i + 1][j] == 1) {
int a = find(i * n + j), b = find((i + 1) * n + j);
if (a != b) {
size[a] += size[b];
}
p[b] = a;
}
if (j < n - 1 && grid[i][j + 1] == 1) {
int a = find(i * n + j), b = find(i * n + j + 1);
if (a != b) {
size[a] += size[b];
}
p[b] = a;
}
}
}
}
int res = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
res = Math.max(res, size[i * n + j]);
}
}
}
return res;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}DFS:
function maxAreaOfIsland(grid: number[][]): number {
let m = grid.length,
n = grid[0].length;
let res = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
res = Math.max(dfs(grid, i, j), res);
}
}
}
return res;
}
function dfs(grid: number[][], i: number, j: number): number {
let m = grid.length,
n = grid[0].length;
if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || grid[i][j] == 0) {
return 0;
}
grid[i][j] = 0;
let res = 1;
for (let [dx, dy] of [
[0, 1],
[0, -1],
[1, 0],
[-1, 0],
]) {
res += dfs(grid, i + dx, j + dy);
}
return res;
}DFS:
class Solution {
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
int res = 0;
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
int t = dfs(grid, i, j, m, n);
res = max(res, t);
}
}
return res;
}
private:
vector<vector<int>> dirs = {{0, 1}, {0, - 1}, {1, 0}, {-1, 0}};
int dfs(vector<vector<int>>& grid, int i, int j, int m, int n) {
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) return 0;
grid[i][j] = 0;
int res = 1;
for (auto dir : dirs)
res += dfs(grid, i + dir[0], j + dir[1], m, n);
return res;
}
};Union find:
class Solution {
public:
vector<int> p;
int maxAreaOfIsland(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> size(m * n, 1);
p.resize(m * n);
for (int i = 0; i < p.size(); ++i) p[i] = i;
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
if (grid[i][j] == 1)
{
if (i < m - 1 && grid[i + 1][j] == 1)
{
int a = find(i * n + j), b = find((i + 1) * n + j);
if (a != b) size[a] += size[b];
p[b] = a;
}
if (j < n - 1 && grid[i][j + 1] == 1)
{
int a = find(i * n + j), b = find(i * n + j + 1);
if (a != b) size[a] += size[b];
p[b] = a;
}
}
}
}
int res = 0;
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
if (grid[i][j] == 1) res = max(res, size[i * n + j]);
}
}
return res;
}
int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};Union find:
var p []int
func maxAreaOfIsland(grid [][]int) int {
m, n := len(grid), len(grid[0])
p = make([]int, m*n)
size := make([]int, m*n)
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
p[i*n+j] = i*n + j
size[i*n+j] = 1
}
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if grid[i][j] == 1 {
if i < m-1 && grid[i+1][j] == 1 {
a, b := find(i*n+j), find((i+1)*n+j)
if a != b {
size[a] += size[b]
}
p[b] = a
}
if j < n-1 && grid[i][j+1] == 1 {
a, b := find(i*n+j), find(i*n+j+1)
if a != b {
size[a] += size[b]
}
p[b] = a
}
}
}
}
res := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if grid[i][j] == 1 {
res = max(res, size[i*n+j])
}
}
}
return res
}
func find(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
func max(a, b int) int {
if a > b {
return a
}
return b
}