題目200. Number of Islands

Given a 2d grid map of '1's (land) and '0's (water), count 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:
11110
11010
11000
00000
Answer: 1
Example 2:
11000
11000
00100
00011
Answer: 3

1,BFS+DFS

public class Solution {
    private int m;
    private int n;
    public int numIslands(char[][] grid) {
        if(grid == null){
            return 0;
        }
        m = grid.length;
        if(m == 0){
            return 0;
        }
        
        n = grid[0].length ;
        if(n == 0){
            return 0;
        }
        
        int nums = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == '1'){
                    dfs(i,j,grid);
                    nums++;
                }
               
            }
        }
        return nums;
    }
    
    private void dfs(int i, int j, char[][] grid){
       if(i < 0 || j < 0 || i >= m || j >= n || grid[i][j] == '0'){
           return;
       }
       grid[i][j] = '0';
       dfs(i-1,j,grid);
       dfs(i+1,j,grid);
       dfs(i,j-1,grid);
       dfs(i,j+1,grid);
    }
}

2,BFS+并查集

public class Solution {
    private int count;
    private int[] parents;
    
    //初始化并查集
    public void initUnionFind(int m, int n, char[][] grid){
        parents = new int[m*n];
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == '1'){
                    count++;
                }
                parents[i*n+j] = i*n+j;
            }
        }
    }
    
    public int find(int idx){
        while(idx != parents[idx]){
            //在查找的過程中壓縮路徑,減少查找的次數
            parents[idx] = parents[parents[idx]];
            idx = parents[idx];
        }
        return idx;
    }
 
    public void union(int p, int q){
        int pRoot = find(p);
        int qRoot = find(q);
        //兩個元素的根不同,則合并
        if(pRoot != qRoot){
            parents[qRoot] = pRoot;
            count--;
        }
    }
    
    public boolean isConnected(int p, int q){
        int pRoot = find(p);
        int qRoot = find(q);
        
        //兩點不連通
        if(pRoot != qRoot){
            return false;
        }
        return false;
    }
    
    
    
    public int numIslands(char[][] grid) {
        if(grid == null || grid.length == 0){
            return 0;
        }
        int m = grid.length;
        
        if(grid[0] == null || grid[0].length == 0){
            return 0;
        }
        int n = grid[0].length;
        
        initUnionFind(m,n,grid);
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == '0'){
                    continue;
                }
                
                int cur = i*n+j;
                if(i > 0 && grid[i-1][j] == '1'){
                    union(cur,cur-n);
                }
                
                if(i < m-1 && grid[i+1][j] == '1'){
                    union(cur,cur+n);
                }
                
                if(j > 0 && grid[i][j-1] == '1'){
                    union(cur, cur-1);
                }
                
                if(j < n-1 && grid[i][j+1] == '1'){
                    union(cur,cur+1);
                }
            }
        }
        return count;
    }
}