200.Number of Islands

       /*
        * 200.Number of Islands 
        * 2016-4-3 by Mingyang
        * union 什么：两个相邻的1元素
     * union 目的：union后计数union集合数量（通过计数union数组中根节点数量）
        */
    class UF {
        public int count = 0;
        public int[] id = null;
        public UF(int m, int n, char[][] grid) {
            for(int i = 0; i < m; i++) {
                for(int j = 0; j < n; j++) {
                    if(grid[i][j] == ‘1‘) count++;
                }
            }
            id = new int[m * n];
            for(int i = 0; i < m * n; i++) {
                id[i] = i;
            }
        }
        public int find(int p) {
            while(p != id[p]) {
                id[p] = id[id[p]];
                p = id[p];
            }
            return p;
        }
        public boolean isConnected(int p, int q) {
            int pRoot = find(p);
            int qRoot = find(q);
            if(pRoot != qRoot) return false;
            else return true;
        }
        public void union(int p, int q) {
            int pRoot = find(p);
            int qRoot = find(q);
            if(pRoot == qRoot) return;
            id[pRoot] = qRoot;
            count--;
        }
        }
    public int numIslands(char[][] grid) {
            if(grid.length == 0 || grid[0].length == 0) return 0;
            int m = grid.length, n = grid[0].length;
            UF uf = new UF(m , n, grid);
            for(int i = 0; i < m; i++) {
                for(int j = 0; j < n; j++) {
                    if(grid[i][j] == ‘0‘) continue;
                    int p = i * n + j;
                    int q;
                    if(i > 0 && grid[i - 1][j] == ‘1‘) {
                        q = p - n;
                        uf.union(p, q);
                    }
                    if(i < m - 1 && grid[i + 1][j] == ‘1‘) {
                        q = p + n;
                        uf.union(p, q);
                    }
                    if(j > 0 && grid[i][j - 1] == ‘1‘) {
                        q = p - 1;
                        uf.union(p, q);
                    }
                    if(j < n - 1 && grid[i][j + 1] == ‘1‘) {
                        q = p + 1;
                        uf.union(p, q);
                    }
                }
            }
            return uf.count;
        }
    //当然你也会觉得下面的可能更简单，那就是另外一种情况了，我就是用下面的方法做的
    //设一个叫count的值，没遇到一个1，就把所有相连的1全部变为0，这样，到底遇到几次1，就是最终有几个小岛啦
       public int numIslands2(char[][] grid) {
           if (grid == null || grid.length == 0 || grid[0].length == 0)
               return 0;
           int count = 0;

           for (int i = 0; i < grid.length; i++) {
               for (int j = 0; j < grid[0].length; j++) {
                   if (grid[i][j] == ‘1‘) {
                       count++;
                       dfs(grid, i, j);
                   }
               }
           }
           return count;
       }
       public void dfs(char[][] grid, int i, int j) {
           // validity checking
           if (i < 0 || j < 0 || i > grid.length - 1 || j > grid[0].length - 1)
               return;
           // if current cell is water or visited
           if (grid[i][j] != ‘1‘)
               return;
           // set visited cell to ‘0‘
           grid[i][j] = ‘0‘;
           // merge all adjacent land
           dfs(grid, i - 1, j);
           dfs(grid, i + 1, j);
           dfs(grid, i, j - 1);
           dfs(grid, i, j + 1);
       }