200. Number of Islands

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200. Number of Islands

Description

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:

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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:

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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'.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    static constexpr int DIRs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

    int numIslands(vector<vector<char>>& 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++) {
                if (grid[i][j] == '1') {
                    queue<pair<int, int>> q;
                    q.push({i, j});
                    grid[i][j] = '0';
                    while (!q.empty()) {
                        auto& [x, y] = q.front();
                        for (int i = 0; i < 4; i++) {
                            int dx = x + DIRs[i][0], dy = y + DIRs[i][1];
                            if (dx >= 0 && dx < m && dy >= 0 && dy < n && grid[dx][dy] == '1') {
                                grid[dx][dy] = '0';
                                q.push({dx, dy});
                            }
                        }
                        q.pop();
                    }
                    res++;
                }
            }
        }
        return res;
    }
};