LeetCode-in-All

200. Number of Islands

Medium

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'.

Solution

void DFS(int row, int col, int m, int n, char** grid) {
    if (row < 0 || row >= m || col < 0 || col >= n || grid[row][col] == '0') {
        return;
    }

    grid[row][col] = '0'; // Mark the current cell as visited

    // Visit the neighboring cells
    DFS(row + 1, col, m, n, grid);
    DFS(row - 1, col, m, n, grid);
    DFS(row, col + 1, m, n, grid);
    DFS(row, col - 1, m, n, grid);
}

int numIslands(char** grid, int gridSize, int* gridColSize) {
    if (gridSize == 0) {
        return 0;
    }

    int m = gridSize;  // ROW
    int n = *gridColSize; // COL
    int numIslands = 0;

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid[i][j] == '1') {  // grid[row][col]
                numIslands++;
                DFS(i, j, m, n, grid);
            }
        }
    }

    return numIslands;
}

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