Number of Islands Solutions in C++

// Source : https://leetcode.com/problems/number-of-islands/
// Author : Hao Chen
// Date   : 2015-06-08



#include <iostream>
#include <vector>
using namespace std;

void mark(vector<vector<char> >& grid, int r, int c){
    if ( r<0 || r>=grid.size() ||
            c<0 || c>=grid[0].size() ||
            grid[r][c] != '1' ) {
        return;
    }

    grid[r][c] = 'M';

    mark(grid, r, c+1); //left
    mark(grid, r, c-1); //right
    mark(grid, r-1, c); //up
    mark(grid, r+1, c); //down
}

int numIslands(vector<vector<char> >& grid) {
    int result = 0;
    for (int r=0; r<grid.size(); r++) {
        for (int c=0; c<grid[r].size(); c++) {
            if (grid[r][c] == '1') {
                result++;
                mark(grid, r, c);
            }
        }
    }
    return result;
}

void initGrid( string g[], int len, vector<vector<char> >& grid )
{
    for (int i=0; i<len; i++){
       grid.push_back(vector<char>(g[i].begin(), g[i].end())); 
    }
}

int main(void)
{
    vector< vector<char> > grid;
    grid.push_back( vector<char>(1, '1') );

    cout << numIslands(grid) << endl;


    grid.clear();

    string g1[] = { "11110",
                    "11010", 
                    "11000", 
                    "00000" };

    initGrid(g1, 4, grid);
    cout << numIslands(grid) << endl;



    grid.clear();

    string g2[] = { "11000",
                    "11000",
                    "00100",
                    "00011" };

    initGrid(g2, 4, grid);
    cout << numIslands(grid) << endl;


    return 0;
}

/// Source : https://leetcode.com/problems/number-of-islands/description/
/// Author : liuyubobobo
/// Time   : 2017-11-18

#include <iostream>
#include <vector>
#include <cassert>

using namespace std;

/// Floodfill - DFS
/// Recursion implementation
///
/// Time Complexity: O(n*m)
/// Space Complexity: O(n*m)
class Solution {

private:
    int d[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int m, n;
    vector<vector<bool>> visited;

    bool inArea(int x, int y){
        return x >= 0 && x < m && y >= 0 && y < n;
    }

    void dfs(vector<vector<char>>& grid, int x, int y){

        //assert(inArea(x,y));
        visited[x][y] = true;
        for(int i = 0; i < 4; i ++){
            int newx = x + d[i][0];
            int newy = y + d[i][1];
            if(inArea(newx, newy) && !visited[newx][newy] && grid[newx][newy] == '1')
                dfs(grid, newx, newy);
        }

        return;
    }

public:
    int numIslands(vector<vector<char>>& grid) {

        m = grid.size();
        if(m == 0)
            return 0;
        n = grid[0].size();
        if(n == 0)
            return 0;

        for(int i = 0 ; i < m ; i ++)
            visited.push_back(vector<bool>(n, false));

        int res = 0;
        for(int i = 0 ; i < m ; i ++)
            for(int j = 0 ; j < n ; j ++)
                if(grid[i][j] == '1' && !visited[i][j]){
                    dfs(grid, i, j);
                    res ++;
                }
        return res;
    }
};


int main() {

    char g1[4][5] = {
            {'1','1','1','1','0'},
            {'1','1','0','1','0'},
            {'1','1','0','0','0'},
            {'0','0','0','0','0'}
    };
    vector<vector<char>> grid1;
    for(int i = 0 ; i < 4 ; i ++)
        grid1.push_back( vector<char>(g1[i], g1[i] + sizeof( g1[i])/sizeof(char)));

    cout << Solution().numIslands(grid1) << endl;
    // 1

    // ---

    char g2[4][5] = {
            {'1','1','0','0','0'},
            {'1','1','0','0','0'},
            {'0','0','1','0','0'},
            {'0','0','0','1','1'}
    };
    vector<vector<char>> grid2;
    for(int i = 0 ; i < 4 ; i ++)
        grid2.push_back(vector<char>(g2[i], g2[i] + sizeof( g2[i])/sizeof(char)));

    cout << Solution().numIslands(grid2) << endl;
    // 2

    return 0;
}

/// Source : https://leetcode.com/problems/number-of-islands/description/
/// Author : liuyubobobo
/// Time   : 2018-08-29

#include <iostream>
#include <vector>
#include <cassert>
#include <stack>

using namespace std;

/// Floodfill - DFS
/// Non-recursion implementation
///
/// Time Complexity: O(n*m)
/// Space Complexity: O(n*m)
class Solution {

private:
    int d[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int m, n;

public:
    int numIslands(vector<vector<char>>& grid) {

        m = grid.size();
        if(m == 0)
            return 0;
        n = grid[0].size();
        if(n == 0)
            return 0;

        vector<vector<bool>> visited(m, vector<bool>(n, false));

        int res = 0;
        for(int i = 0 ; i < m ; i ++)
            for(int j = 0 ; j < n ; j ++)
                if(grid[i][j] == '1' && !visited[i][j]){
                    dfs(grid, i, j, visited);
                    res ++;
                }
        return res;
    }

private:
    void dfs(vector<vector<char>>& grid, int x, int y, vector<vector<bool>>& visited){

        stack<pair<int, int>> q;
        q.push(make_pair(x, y));
        visited[x][y] = true;
        while(!q.empty()){
            int curx = q.top().first;
            int cury = q.top().second;
            q.pop();

            for(int i = 0; i < 4; i ++){
                int newX = curx + d[i][0];
                int newY = cury + d[i][1];
                if(inArea(newX, newY) && !visited[newX][newY] && grid[newX][newY] == '1'){
                    q.push(make_pair(newX, newY));
                    visited[newX][newY] = true;
                }
            }
        }

        return;
    }

    bool inArea(int x, int y){
        return x >= 0 && x < m && y >= 0 && y < n;
    }
};


int main() {

    vector<vector<char>> grid1 = {
            {'1','1','1','1','0'},
            {'1','1','0','1','0'},
            {'1','1','0','0','0'},
            {'0','0','0','0','0'}
    };
    cout << Solution().numIslands(grid1) << endl;
    // 1

    // ---

    vector<vector<char>> grid2 = {
            {'1','1','0','0','0'},
            {'1','1','0','0','0'},
            {'0','0','1','0','0'},
            {'0','0','0','1','1'}
    };
    cout << Solution().numIslands(grid2) << endl;
    // 3

    return 0;
}

/// Source : https://leetcode.com/problems/number-of-islands/description/
/// Author : liuyubobobo
/// Time   : 2018-08-25

#include <iostream>
#include <vector>
#include <cassert>
#include <queue>

using namespace std;

/// Floodfill - BFS
/// Time Complexity: O(n*m)
/// Space Complexity: O(n*m)
class Solution {

private:
    int d[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int m, n;

public:
    int numIslands(vector<vector<char>>& grid) {

        m = grid.size();
        if(m == 0)
            return 0;
        n = grid[0].size();
        if(n == 0)
            return 0;

        vector<vector<bool>> visited(m, vector<bool>(n, false));

        int res = 0;
        for(int i = 0 ; i < m ; i ++)
            for(int j = 0 ; j < n ; j ++)
                if(grid[i][j] == '1' && !visited[i][j]){
                    bfs(grid, i, j, visited);
                    res ++;
                }
        return res;
    }

private:
    void bfs(vector<vector<char>>& grid, int x, int y, vector<vector<bool>>& visited){

        queue<pair<int, int>> q;
        q.push(make_pair(x, y));
        visited[x][y] = true;
        while(!q.empty()){
            int curx = q.front().first;
            int cury = q.front().second;
            q.pop();

            for(int i = 0; i < 4; i ++){
                int newX = curx + d[i][0];
                int newY = cury + d[i][1];
                if(inArea(newX, newY) && !visited[newX][newY] && grid[newX][newY] == '1'){
                    q.push(make_pair(newX, newY));
                    visited[newX][newY] = true;
                }
            }
        }

        return;
    }

    bool inArea(int x, int y){
        return x >= 0 && x < m && y >= 0 && y < n;
    }
};


int main() {

    vector<vector<char>> grid1 = {
            {'1','1','1','1','0'},
            {'1','1','0','1','0'},
            {'1','1','0','0','0'},
            {'0','0','0','0','0'}
    };
    cout << Solution().numIslands(grid1) << endl;
    // 1

    // ---

    vector<vector<char>> grid2 = {
            {'1','1','0','0','0'},
            {'1','1','0','0','0'},
            {'0','0','1','0','0'},
            {'0','0','0','1','1'}
    };
    cout << Solution().numIslands(grid2) << endl;
    // 3

    return 0;
}

/// Source : https://leetcode.com/problems/number-of-islands/description/
/// Author : liuyubobobo
/// Time   : 2018-08-26

#include <iostream>
#include <vector>
#include <cassert>
#include <unordered_set>

using namespace std;

/// Using union-find
/// Time Complexity: O(n*m)
/// Space Complexity: O(n*m)
class UnionFind{

private:
    vector<int> rank, parent;

public:
    UnionFind(int n){
        rank.clear();
        parent.clear();
        for( int i = 0 ; i < n ; i ++ ){
            parent.push_back(i);
            rank.push_back(1);
        }
    }

    int find(int p){
        while(p != parent[p]){
            parent[p] = parent[parent[p]];
            p = parent[p];
        }
        return p;
    }

    bool isConnected(int p , int q){
        return find(p) == find(q);
    }

    void unionElements(int p, int q){

        int pRoot = find(p);
        int qRoot = find(q);

        if(pRoot == qRoot)
            return;

        if(rank[pRoot] < rank[qRoot])
            parent[pRoot] = qRoot;
        else if(rank[qRoot] < rank[pRoot])
            parent[qRoot] = pRoot;
        else{ // rank[pRoot] == rank[qRoot]
            parent[pRoot] = qRoot;
            rank[qRoot] += 1;
        }
    }
};

class Solution {

private:
    int d[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int m, n;

public:
    int numIslands(vector<vector<char>>& grid) {

        m = grid.size();
        if(m == 0)
            return 0;
        n = grid[0].size();
        if(n == 0)
            return 0;

        UnionFind uf(m * n);
        for(int i = 0; i < m; i ++)
            for(int j = 0; j < n; j ++)
                if(grid[i][j] == '1')
                    for(int k = 0; k < 4; k ++){
                        int newX = i + d[k][0];
                        int newY = j + d[k][1];
                        if(inArea(newX, newY) && grid[newX][newY] == '1')
                            uf.unionElements(i * n + j, newX * n + newY);
                    }

        unordered_set<int> regions;
        for(int i = 0 ; i < m; i ++)
            for(int j = 0; j < n; j ++)
                if(grid[i][j] == '1')
                    regions.insert(uf.find(i * n + j));
        return regions.size();
    }

private:
    bool inArea(int x, int y){
        return x >= 0 && x < m && y >= 0 && y < n;
    }
};


int main() {

    vector<vector<char>> grid1 = {
            {'1','1','1','1','0'},
            {'1','1','0','1','0'},
            {'1','1','0','0','0'},
            {'0','0','0','0','0'}
    };
    cout << Solution().numIslands(grid1) << endl;
    // 1

    // ---

    vector<vector<char>> grid2 = {
            {'1','1','0','0','0'},
            {'1','1','0','0','0'},
            {'0','0','1','0','0'},
            {'0','0','0','1','1'}
    };
    cout << Solution().numIslands(grid2) << endl;
    // 3

    return 0;
}