This approach uses Depth-First Search (DFS) to explore the grid. We iterate over each cell in the grid, and every time we find an unvisited '1', it indicates the discovery of a new island. We then perform DFS from that cell to mark all the connected '1's as visited, effectively marking the entire island.
Time Complexity: O(m*n) where m is the number of rows and n is the number of columns since each cell is visited once.
Space Complexity: O(m*n) in the worst case due to the recursion stack used by DFS.
1#include <stdbool.h>
2
3void dfs(char** grid, int r, int c, int gridSize, int* gridColSize) {
4 if (r < 0 || c < 0 || r >= gridSize || c >= gridColSize[r] || grid[r][c] == '0') {
5 return;
6 }
7 grid[r][c] = '0';
8 dfs(grid, r-1, c, gridSize, gridColSize);
9 dfs(grid, r+1, c, gridSize, gridColSize);
10 dfs(grid, r, c-1, gridSize, gridColSize);
11 dfs(grid, r, c+1, gridSize, gridColSize);
12}
13
14int numIslands(char** grid, int gridSize, int* gridColSize) {
15 int numIslands = 0;
16 for (int i = 0; i < gridSize; i++) {
17 for (int j = 0; j < gridColSize[i]; j++) {
18 if (grid[i][j] == '1') {
19 numIslands++;
20 dfs(grid, i, j, gridSize, gridColSize);
21 }
22 }
23 }
24 return numIslands;
25}The function dfs is responsible for marking connected '1's as visited by changing them to '0'. The numIslands function traverses each cell in the grid, and upon finding a '1', it increments the count and triggers a DFS from that cell to mark the entire island.
This approach uses Breadth-First Search (BFS) to traverse the grid. Similar to DFS, we treat each '1' as a node in a graph. On encountering a '1', we initiate a BFS to explore all connected '1's (island nodes) by utilizing a queue for the frontier, marking them in-place as visited.
Time Complexity: O(m*n), where m and n are rows and columns.#
Space Complexity: O(min(m, n)) due to the queue storage in BFS.
1import java.util.LinkedList;
2import java.util.Queue;
3
4public class Solution {
5 private static final int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
6
7 public int numIslands(char[][] grid) {
8 if (grid == null || grid.length == 0) {
9 return 0;
10 }
11 int numIslands = 0;
12 for (int r = 0; r < grid.length; r++) {
13 for (int c = 0; c < grid[0].length; c++) {
14 if (grid[r][c] == '1') {
15 numIslands++;
16 grid[r][c] = '0';
17 Queue<int[]> queue = new LinkedList<>();
18 queue.add(new int[]{r, c});
19 while (!queue.isEmpty()) {
20 int[] cell = queue.poll();
21 for (int[] d : directions) {
22 int nr = cell[0] + d[0], nc = cell[1] + d[1];
23 if (nr >= 0 && nr < grid.length && nc >= 0 && nc < grid[0].length && grid[nr][nc] == '1') {
24 queue.add(new int[]{nr, nc});
25 grid[nr][nc] = '0';
26 }
27 }
28 }
29 }
30 }
31 }
32 return numIslands;
33 }
34}Java's solution leverages a Queue to implement BFS. When encountering '1', BFS starts and expands, visiting the whole component marking each cell and exploring adjacent '1's iteratively using the queue.