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‎README.md

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@@ -408,7 +408,7 @@ Your ideas/fixes/algorithms are more than welcome!
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|203|[Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_203.java)| O(n)|O(1) | Easy
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|202|[Happy Number](https://leetcode.com/problems/happy-number/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_202.java)| O(k)|O(k) | Easy
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|201|[Bitwise AND of Numbers Range](https://leetcode.com/problems/bitwise-and-of-numbers-range/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_201.java)| O(min(m,n))|O(1) | Medium | Bit Manipulation
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|200|[Number of Islands](https://leetcode.com/problems/number-of-islands/)|[Union Find](../master/src/main/java/com/fishercoder/solutions/_200UnionFind.java) [DFS](../master/MEDIUM/src/medium/_200DFS.java)| O(m*n)|O(m*n) | Medium| Union Find, DFS
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|200|[Number of Islands](https://leetcode.com/problems/number-of-islands/)|[Solution](../master/MEDIUM/src/medium/_200.java)| O(m*n)|O(m*n) | Medium| Union Find, DFS
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|199|[Binary Tree Right Side View](https://leetcode.com/problems/binary-tree-right-side-view/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_199.java)| O(n)|O(n)| Medium | BFS
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|198|[House Robber](https://leetcode.com/problems/house-robber/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_198.java)| O(n)|O(n)| Easy | DP
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|191|[Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_191.java)| O(n)|O(1)| Easy | Bit Manipulation

‎src/main/java/com/fishercoder/solutions/_200.java

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package com.fishercoder.solutions;
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/**
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* 200. Number of Islands
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* Given a 2d grid map of '1's (land) and '0's (water),
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* count the number of islands.
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* An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically.
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* You may assume all four edges of the grid are all surrounded by water.
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Example 1:
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11110
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11010
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11000
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00000
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Answer: 1
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Example 2:
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11000
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11000
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00100
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00011
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Answer: 3
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*/
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public class _200 {
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public static class DFSSolution {
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public int numIslands(char[][] grid) {
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if (grid == null || grid.length == 0) return 0;
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int count = 0;
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int m = grid.length, n = grid[0].length;
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (grid[i][j] == '1') {
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count++;
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dfs(grid, i, j, m, n);
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}
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}
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}
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return count;
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}
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void dfs(char[][] grid, int i, int j, int m, int n) {
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if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0') return;
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grid[i][j] = '0';
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dfs(grid, i + 1, j, m, n);
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dfs(grid, i, j + 1, m, n);
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dfs(grid, i - 1, j, m, n);
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dfs(grid, i, j - 1, m, n);
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}
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}
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public static class UnionFindSolution {
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class UnionFind{
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int count;
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int m, n;
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int[] ids;
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public UnionFind(char[][] grid){
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m = grid.length;
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n = grid[0].length;
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ids = new int[m*n];
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for(int i = 0; i < m; i++){
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for(int j = 0; j < n; j++){
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if(grid[i][j] == '1') {
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count++;
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ids[i*n+j] = i*n+j;
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}
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}
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}
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}
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public void union(int i, int j){
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int x = find(ids, i);
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int y = find(ids, j);
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if(x != y) {/**note: this is when x != y,
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only in this case, we should union these two nodes, which makes sense naturally.*/
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count--;
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ids[x] = y;//ids[y] = x; //also works
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}
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}
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public int find(int[] ids, int i){
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if(ids[i] == i) return i;
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return find(ids, ids[i]);
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}
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}
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public int numIslands(char[][] grid) {
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if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
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int[] dirs = new int[]{0,1,0,-1,0};
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UnionFind uf = new UnionFind(grid);
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int m = grid.length, n = grid[0].length;
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for(int i = 0; i < m; i++){
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for(int j = 0; j < n; j++){
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if(grid[i][j] == '1'){
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for(int k = 0; k < 4; k++){
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int x = i+dirs[k];
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int y = j+dirs[k+1];
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if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1'){
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int id1 = i*n+j;
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int id2 = x*n+y;
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uf.union(id1, id2);
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}
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}
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}
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}
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}
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return uf.count;
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}
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}
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}

‎src/main/java/com/fishercoder/solutions/_200DFS.java

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‎src/main/java/com/fishercoder/solutions/_200UnionFind.java

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‎src/test/java/com/fishercoder/_200Test.java

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package com.fishercoder;
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import com.fishercoder.solutions._200;
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import org.junit.BeforeClass;
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import org.junit.Test;
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import static org.junit.Assert.assertEquals;
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public class _200Test {
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private static _200.DFSSolution dfsSolution;
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private static _200.UnionFindSolution unionFindSolution;
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private static char[][] grid;
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@BeforeClass
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public static void setup(){
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dfsSolution = new _200.DFSSolution();
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unionFindSolution = new _200.UnionFindSolution();
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}
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@Test
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public void test1(){
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grid = new char[][]{
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{'1','1','1'},
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{'0','1','0'},
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{'1','1','1'},
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};
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assertEquals(1, dfsSolution.numIslands(grid));
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assertEquals(1, unionFindSolution.numIslands(grid));
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}
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@Test
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public void test2(){
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grid = new char[][]{
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{'1','1','1','1','0'},
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{'1','1','0','1','0'},
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{'1','1','0','0','0'},
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{'0','0','0','0','0'},
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};
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assertEquals(0, dfsSolution.numIslands(grid));
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assertEquals(0, unionFindSolution.numIslands(grid));
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}
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@Test
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public void test3(){
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grid = new char[][]{
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{'1','1','0','0','0'},
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{'1','1','0','0','0'},
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{'0','0','1','0','0'},
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{'0','0','0','1','1'},
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};
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assertEquals(3, dfsSolution.numIslands(grid));
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assertEquals(3, unionFindSolution.numIslands(grid));
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}
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}

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