/** 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 checkingif (i < 0 || j < 0 || i > grid.length - 1 || j > grid[0].length - 1)return;// if current cell is water or visitedif (grid[i][j] != '1')return;// set visited cell to '0'grid[i][j] = '0';// merge all adjacent landdfs(grid, i - 1, j);dfs(grid, i + 1, j);dfs(grid, i, j - 1);dfs(grid, i, j + 1);}