leetcode 695. Max Area of Island

题目解析：找到最大连通块。并查集模板题。

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) 
    {
        if(grid.empty()) return 0 ;
        int m = grid.size() , n = grid[0].size() ;
        
        root = vector<int>(m * n , 0) ;
        cnt = vector<int>(m * n , 0) ;
        int cnt1 = 0 ;
        
        for(int i = 0 ; i < m ; i++)
            for(int j = 0 ; j < n ; j++)
            {
                root[i * n + j] = i * n + j ;
                
                if(grid[i][j] == 1)
                {
                    cnt[i * n + j] = 1 ;
                    cnt1++ ;
                }
            }
        
        if(cnt1 == 0) return 0 ;
        
        int res = INT_MIN ;
        for(int i = 0 ; i < m ; i++)
            for(int j = 0 ; j < n ; j++)
            {
                if(grid[i][j] == 0) continue ;
                
                res = max(res , cnt[find(i*n+j)]) ;
                for(int k = 0 ; k < 4 ; k++)
                {
                    int ni = i + dir[k] , nj = j + dir[k + 1] ;
                    if(ni < 0 || ni >= m || nj < 0 || nj >= n || grid[ni][nj] == 0) continue ;
                    
                    int x = find(i * n + j) , y = find(ni * n + nj) ;
                    if(x != y)
                    {
                        root[x] = y ;
                        cnt[y] += cnt[x] ;
                        res = max(res , cnt[y]) ;
                    }
                }
            }
        
        return res ;
    }
    
private:
    
    vector<int> root , cnt , dir = {-1 , 0 , 1 , 0 , -1};
    
    int find(int x)
    {
        return root[x] == x ? x : root[x] = find(root[x]) ;
    }
};

dfs解法：其实也是查找连通块的另一种解法

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) 
    {
        if(grid.empty()) return 0 ;
        m = grid.size() ;
        n = grid[0].size() ;
        
        vis = vector<int>(m * n , 0) ;
        int res = 0 ;
        
        for(int i = 0 ; i < m ; i++)
            for(int j = 0 ; j < n ; j++)
            {
                if(grid[i][j] == 0 || vis[i * n + j]) continue ;
                
                int cnt = 0 ;
                dfs(i , j , cnt, grid) ;
                res = max(res , cnt) ;
            }
                  
        return res ;
    }
    
private:
    
    vector<int> vis , dir = {-1 , 0 , 1 , 0 , -1};
    int m , n ;
    
    void dfs(int i , int j , int& cnt , vector<vector<int>>& grid)
    {
        cnt++ ;
        vis[i * n + j] = 1 ;
        
        for(int k = 0 ; k < 4 ; k++)
        {
            int ni = i + dir[k] , nj = j + dir[k + 1] ;
            if(ni < 0 || ni >= m || nj < 0 || nj >= n || grid[ni][nj] == 0 || vis[ni * n + nj]) continue ;
            dfs(ni , nj , cnt , grid) ;
        }
    }
};

　　dfs里为了避免陷入循环，在dfs之前应该要将vis值置1

标签：ni,cnt,Area,int,Max,nj,++,grid,leetcode
来源： https://www.cnblogs.com/mychen06/p/12636486.html