
 
 其实还是用并查集将 独立的岛屿视为独立的子集。
 count其实是集合的个数,同一个块岛屿被压缩成了一个集合,而每个表示海洋的格子依然被看作独立的集合,在所有的格子都走完一遍后,count = 被压缩的岛屿 + 所有表示海洋的独立格子的数目,因此减去格子的数目比较重要。
class Solution {
    private int rows;
    private int cols;
    public int numIslands(char[][] grid) {
        rows = grid.length;
        if (rows == 0) {
            return 0;
        }    
        cols = grid[0].length;
        //空地的数量
        int spaces = 0;
        UnionFind unionFind = new UnionFind(cols * rows);
        int[][] directions = {{1, 0}, {0, 1}};
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '0') {
                    spaces++;
                } else {
                    for (int[] direction : directions) {
                        int newX = i + direction[0];
                        int newY = j + direction[1];
                        if (newX < rows && newY < cols && grid[newX][newY] == '1') {
                            unionFind.union(getIndex(i, j), getIndex(newX, newY));
                        }
                    }
                }
            }
        }
        //没懂为什么要减去spaces
        return unionFind.getCount() - spaces;
    }
    private int getIndex(int i, int j) {
        return i * cols + j;
    }
    private class UnionFind{
        private int count;
        private int[] parent;
        public int getCount() {
            return count;
        }
        public UnionFind(int n) {
            this.count = n;
            parent = new int[n];
            for (int i = 0;  i < n; i++) {
                parent[i] = i;
            }
        }
        private int find(int x) {
            while (x != parent[x]) {
                parent[x] = parent[parent[x]];
                x = parent[x];
            }
            return x;
        }
        public void union(int x, int y) {
            int xRoot = find(x);
            int yRoot = find(y);
            if (xRoot == yRoot) {
                return;
            }
            parent[xRoot] = yRoot;
            count--;
        }
    }
}



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