LeetCode200岛屿数量（DFS+BFS）C++实现

https://leetcode-cn.com/problems/number-of-islands/

深度优先搜索(DFS)

class Solution {
public:
    //时间复杂度：O(MN), 其中M和N分别为行数和列数。
    //空间复杂度：O(MN), 在最坏情况下，整个网格均为陆地，深度优先搜索的深度达到MN。
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        if(!m) return 0;
        int n = grid[0].size();
        int count = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == '1'){
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }
private:
    void dfs(vector<vector<char>>& grid, int i, int j) {
        int m = grid.size();
        int n = grid[0].size();
        grid[i][j] = '0';
        if(i - 1>= 0 && grid[i-1][j] == '1') dfs(grid, i - 1, j);
        if(i + 1 < m && grid[i+1][j] == '1') dfs(grid, i + 1, j);
        if(j - 1 >= 0 && grid[i][j-1] == '1') dfs(grid, i, j - 1);
        if(j + 1 < n && grid[i][j+1] == '1') dfs(grid, i, j + 1);
    }
};

广度优先搜索(BFS)

class Solution {
public:
    //时间复杂度：O(MN)，其中M和N分别为行数和列数。
    //空间复杂度：O(min⁡(M,N))，在最坏情况下，整个网格均为陆地，队列的大小可以达到 min⁡(M,N)。
    int numIslands(vector<vector<char>>& grid) {
        int nr = grid.size();
        if (!nr) return 0;
        int nc = grid[0].size();

        int num_islands = 0;
        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    ++num_islands;
                    grid[r][c] = '0';
                    queue<pair<int, int>> neighbors;
                    neighbors.push({r, c});
                    while (!neighbors.empty()) {
                        auto rc = neighbors.front();
                        neighbors.pop();
                        int row = rc.first, col = rc.second;
                        if (row - 1 >= 0 && grid[row-1][col] == '1') {
                            neighbors.push({row-1, col});
                            grid[row-1][col] = '0';
                        }
                        if (row + 1 < nr && grid[row+1][col] == '1') {
                            neighbors.push({row+1, col});
                            grid[row+1][col] = '0';
                        }
                        if (col - 1 >= 0 && grid[row][col-1] == '1') {
                            neighbors.push({row, col-1});
                            grid[row][col-1] = '0';
                        }
                        if (col + 1 < nc && grid[row][col+1] == '1') {
                            neighbors.push({row, col+1});
                            grid[row][col+1] = '0';
                        }
                    }
                }
            }
        }

        return num_islands;
    }
};