LeetCode //C - 200. Number of Islands

200. Number of Islands

Given an m x n 2D binary grid grid which represents a map of *‘1’*s (land) and *‘0’*s (water), return 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: grid = [
[“1”,“1”,“1”,“1”,“0”],
[“1”,“1”,“0”,“1”,“0”],
[“1”,“1”,“0”,“0”,“0”],
[“0”,“0”,“0”,“0”,“0”]
]
Output: 1

Example 2:

Input: grid = [
[“1”,“1”,“0”,“0”,“0”],
[“1”,“1”,“0”,“0”,“0”],
[“0”,“0”,“1”,“0”,“0”],
[“0”,“0”,“0”,“1”,“1”]
]
Output: 3

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is ‘0’ or ‘1’.

From: LeetCode
Link: 200. Number of Islands

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Solution:

Ideas：

1. Initialize: The numIslands function initializes the count of islands to zero. It also determines the dimensions of the grid, m (number of rows) and n (number of columns).

2. Iterate through the Grid: The code then iterates over each cell of the grid. If the current cell is ‘1’, it indicates a part of an island that hasn’t been explored yet.

3. DFS to Explore the Island: When a ‘1’ is found, we use the dfs function to explore the entire island. The idea is that once we find a ‘1’, we want to explore all its neighboring cells (up, down, left, and right) to see if they are also part of the same island. If they are, we continue the exploration recursively.

• In the dfs function, if the current cell is out-of-bounds, or if it is water (i.e., ‘0’), we return immediately without further exploration.
• If the current cell is ‘1’, we mark it as visited by changing its value to ‘0’. This ensures that we don’t count the same part of an island more than once.
• We then recursively call dfs on the neighboring cells to continue exploring the island.

4. Counting the Islands: Each time we initiate a DFS exploration from the numIslands function (i.e., every time we find an unexplored ‘1’), we increment our island count by 1. By the end of the iteration over the grid, we would have explored and counted all distinct islands.

5. Return the Count: Finally, numIslands returns the count of islands.

Code:

void dfs(char** grid, int i, int j, int m, int n) {if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') {return;}grid[i][j] = '0';dfs(grid, i - 1, j, m, n); // updfs(grid, i + 1, j, m, n); // downdfs(grid, i, j - 1, m, n); // leftdfs(grid, i, j + 1, m, n); // right
}int numIslands(char** grid, int gridSize, int* gridColSize) {if (grid == NULL || gridSize == 0 || gridColSize == NULL) {return 0;}int m = gridSize;int n = gridColSize[0];int islandCount = 0;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == '1') {islandCount++;dfs(grid, i, j, m, n);}}}return islandCount;
}