算法-16-并查集

并查集简介 

并查集：一开始，把a，b，c放入并查集，a自己一个集合，b自己一个，c自己一个

提供的方法        1.boolean isSameSet(a,b)，判断ab是否在同一个集合

                          2.void union(a,b),把a所在集合的全部，和b所在集合的全部，两个大集合合并

怎么实现的方法一

对于每个元素，都有一个指针指向自己

 查询a，b是否在同一个集合，对于a，a往上到不能再往上的节点就是他的代表节点，也就是a

b同理，代表节点不同，所以ab所在的集合不同

我们把一个节点往上找，找到不能再往上的那个节点叫集合的代表节点

那如何实现union合并呢

并查集有自己的机制，记录每一个集合大小是多少

然后union（a，e）时，把b的指针直接挂到a后面，union（a，c）时，小的挂到大的上，c挂到a后面

如果再union（b,d）然后union(d,e)怎么union，先顺着d往上找找到b，然后顺着e往上找找到a，小的挂大的，只需改代表节点b->a,欧了

import java.util.HashMap;
import java.util.List;
import java.util.Stack;public class Code05_UnionFind {public static class Node<V> {V value;public Node(V v) {value = v;}}public static class UnionFind<V> {//搞三张表public HashMap<V, Node<V>> nodes;public HashMap<Node<V>, Node<V>> parents;//parent表，key是儿子，value是父亲，儿子父亲一一对应public HashMap<Node<V>, Integer> sizeMap;//代表节点和大小size对应public UnionFind(List<V> values) {//初始化nodes = new HashMap<>();parents = new HashMap<>();sizeMap = new HashMap<>();for (V cur : values) {Node<V> node = new Node<>(cur);nodes.put(cur, node);parents.put(node, node);sizeMap.put(node, 1);}}// 给你一个节点，请你往上到不能再往上，把代表返回//找代表节点要用很多次，所以把一次找一个爹优化成直接找祖宗，沿一条链压入栈，//一个一个弹出，只要不是栈顶就挂到祖宗后面public Node<V> findFather(Node<V> cur) {Stack<Node<V>> path = new Stack<>();while (cur != parents.get(cur)) {path.push(cur);cur = parents.get(cur);}//推出的时候cur = parents.get(cur）while (!path.isEmpty()) {parents.put(path.pop(), cur);}return cur;}public boolean isSameSet(V a, V b) {return findFather(nodes.get(a)) == findFather(nodes.get(b));}public void union(V a, V b) {Node<V> aHead = findFather(nodes.get(a));Node<V> bHead = findFather(nodes.get(b));if (aHead != bHead) {int aSetSize = sizeMap.get(aHead);int bSetSize = sizeMap.get(bHead);Node<V> big = aSetSize >= bSetSize ? aHead : bHead;//就是if（a>b），big=a，small=bNode<V> small = big == aHead ? bHead : aHead;parents.put(small, big);//小集合的头部（父亲）设成大集合sizeMap.put(big, aSetSize + bSetSize);//大集合变大sizeMap.remove(small);}}public int sets() {return sizeMap.size();}}
}

并查集应用

1.表示任务关系，1代表认识，0代表不认识

一定是正方形N*N，且m[i][j]=1,那么m[j][i]=1,大正方形是对称的

求有多少个联通区（朋友圈，互相认识的域）

用并查集，从0开始，0认识2认识4，一合并【0.2.4】，从1开始【1.3】一共两个朋友圈

// 本题为leetcode原题
// 测试链接：https://leetcode.com/problems/friend-circles/
// 可以直接通过
public class Code01_FriendCircles {public static int findCircleNum(int[][] M) {int N = M.length;// 初始化{0} {1} {2} {N-1}每个数据单独成一个集合UnionFind unionFind = new UnionFind(N);for (int i = 0; i < N; i++) {	//行for (int j = i + 1; j < N; j++) {//列，for循环只用便利右上部分，因为1认识2，2就认识1if (M[i][j] == 1) { // i和j互相认识unionFind.union(i, j);//i，j合并}}}return unionFind.sets();}public static class UnionFind {//用数组代替哈希表，更快// parent[i] = k ： i的父亲是kprivate int[] parent;// size[i] = k ： 如果i是代表节点，size[i]才有意义，否则无意义// i所在的集合大小是多少private int[] size;// 辅助结构，作栈用private int[] help;// 一共有多少个集合private int sets;//初始化public UnionFind(int N) {parent = new int[N];size = new int[N];help = new int[N];sets = N;for (int i = 0; i < N; i++) {parent[i] = i;	//i的父亲节点就是自己size[i] = 1;	//i自己就是代表节点，大小为1}}// 从i开始一直往上，往上到不能再往上，代表节点，返回// 这个过程要做路径压缩private int find(int i) {int hi = 0;while (i != parent[i]) {//i不等于自己的父亲，i往上 help[hi++] = i;//help作栈i = parent[i];}//直到i=parent[i]也就是代表节点了for (hi--; hi >= 0; hi--) {parent[help[hi]] = i;}return i;}public void union(int i, int j) {int f1 = find(i);int f2 = find(j);if (f1 != f2) {if (size[f1] >= size[f2]) {size[f1] += size[f2];parent[f2] = f1;} else {size[f2] += size[f1];parent[f1] = f2;}sets--;}}public int sets() {return sets;}}}

岛问题

给定一个二维数组matrix，里面的值不是1就是0

上或下或左或右相邻的1认为是一片岛

返回matrix中岛的数量，求几片1

思路，从左往右依次遍历，不是1就跳过，是一就调用infection函数，把与这个一相邻的1全部变成2，然后往后走不是1就跳

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Stack;// 本题为leetcode原题
// 测试链接：https://leetcode.com/problems/number-of-islands/
// 所有方法都可以直接通过
public class Code02_NumberOfIslands {public static int numIslands3(char[][] board) {int islands = 0;for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[0].length; j++) {if (board[i][j] == '1') {//遍历，是1岛的数量++然后调用感染函数islands++;infect(board, i, j);}}}return islands;}// 从(i,j)这个位置出发，把所有练成一片的'1'字符，变成0，不是0字符而是0阿斯克码值public static void infect(char[][] board, int i, int j) {if (i < 0 || i == board.length || j < 0 || j == board[0].length || board[i][j] != '1') {return;}board[i][j] = 0;//一定要改，否则无限递归infect(board, i - 1, j);infect(board, i + 1, j);infect(board, i, j - 1);infect(board, i, j + 1);}public static int numIslands1(char[][] board) {int row = board.length;int col = board[0].length;Dot[][] dots = new Dot[row][col];List<Dot> dotList = new ArrayList<>();for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {if (board[i][j] == '1') {dots[i][j] = new Dot();dotList.add(dots[i][j]);}}}UnionFind1<Dot> uf = new UnionFind1<>(dotList);for (int j = 1; j < col; j++) {//第零行没有上，单独拿出来// (0,j)  (0,0)跳过了  (0,1) (0,2) (0,3)if (board[0][j - 1] == '1' && board[0][j] == '1') {//自己是1，右边也是1就合并uf.union(dots[0][j - 1], dots[0][j]);}}for (int i = 1; i < row; i++) {{//第零列没有左，单独拿出来if (board[i - 1][0] == '1' && board[i][0] == '1') {uf.union(dots[i - 1][0], dots[i][0]);}}for (int i = 1; i < row; i++) {//既有左又有上for (int j = 1; j < col; j++) {if (board[i][j] == '1') {if (board[i][j - 1] == '1') {uf.union(dots[i][j - 1], dots[i][j]);}if (board[i - 1][j] == '1') {uf.union(dots[i - 1][j], dots[i][j]);}}}}return uf.sets();}public static class Dot {}public static class Node<V> {V value;public Node(V v) {value = v;}}public static class UnionFind1<V> {public HashMap<V, Node<V>> nodes;public HashMap<Node<V>, Node<V>> parents;public HashMap<Node<V>, Integer> sizeMap;public UnionFind1(List<V> values) {nodes = new HashMap<>();parents = new HashMap<>();sizeMap = new HashMap<>();for (V cur : values) {Node<V> node = new Node<>(cur);nodes.put(cur, node);parents.put(node, node);sizeMap.put(node, 1);}}public Node<V> findFather(Node<V> cur) {Stack<Node<V>> path = new Stack<>();while (cur != parents.get(cur)) {path.push(cur);cur = parents.get(cur);}while (!path.isEmpty()) {parents.put(path.pop(), cur);}return cur;}public void union(V a, V b) {Node<V> aHead = findFather(nodes.get(a));Node<V> bHead = findFather(nodes.get(b));if (aHead != bHead) {int aSetSize = sizeMap.get(aHead);int bSetSize = sizeMap.get(bHead);Node<V> big = aSetSize >= bSetSize ? aHead : bHead;Node<V> small = big == aHead ? bHead : aHead;parents.put(small, big);sizeMap.put(big, aSetSize + bSetSize);sizeMap.remove(small);}}public int sets() {return sizeMap.size();}}public static int numIslands2(char[][] board) {int row = board.length;int col = board[0].length;UnionFind2 uf = new UnionFind2(board);for (int j = 1; j < col; j++) {if (board[0][j - 1] == '1' && board[0][j] == '1') {uf.union(0, j - 1, 0, j);}}for (int i = 1; i < row; i++) {if (board[i - 1][0] == '1' && board[i][0] == '1') {uf.union(i - 1, 0, i, 0);}}for (int i = 1; i < row; i++) {for (int j = 1; j < col; j++) {if (board[i][j] == '1') {if (board[i][j - 1] == '1') {uf.union(i, j - 1, i, j);}if (board[i - 1][j] == '1') {uf.union(i - 1, j, i, j);}}}}return uf.sets();}public static class UnionFind2 {//把二维数组转成一维，对二维数组（i，j）->i*列+jprivate int[] parent;private int[] size;private int[] help;private int col;//列private int sets;//集合数量public UnionFind2(char[][] board) {col = board[0].length;sets = 0;int row = board.length;int len = row * col;//一共有几个数parent = new int[len];size = new int[len];help = new int[len];for (int r = 0; r < row; r++) {//初始化for (int c = 0; c < col; c++) {if (board[r][c] == '1') {int i = index(r, c);parent[i] = i;size[i] = 1;sets++;}}}}// (r,c) -> i，用i代替（r，c）private int index(int r, int c) {//r行c列对应的下标return r * col + c;}// 原始位置 -> 下标private int find(int i) {int hi = 0;while (i != parent[i]) {help[hi++] = i;i = parent[i];}for (hi--; hi >= 0; hi--) {parent[help[hi]] = i;}return i;}public void union(int r1, int c1, int r2, int c2) {int i1 = index(r1, c1);int i2 = index(r2, c2);int f1 = find(i1);int f2 = find(i2);if (f1 != f2) {if (size[f1] >= size[f2]) {size[f1] += size[f2];parent[f2] = f1;} else {size[f2] += size[f1];parent[f1] = f2;}sets--;}}public int sets() {return sets;}}// 为了测试public static char[][] generateRandomMatrix(int row, int col) {char[][] board = new char[row][col];for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {board[i][j] = Math.random() < 0.5 ? '1' : '0';}}return board;}// 为了测试public static char[][] copy(char[][] board) {int row = board.length;int col = board[0].length;char[][] ans = new char[row][col];for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {ans[i][j] = board[i][j];}}return ans;}// 为了测试public static void main(String[] args) {int row = 0;int col = 0;char[][] board1 = null;char[][] board2 = null;char[][] board3 = null;long start = 0;long end = 0;row = 1000;col = 1000;board1 = generateRandomMatrix(row, col);board2 = copy(board1);board3 = copy(board1);System.out.println("感染方法、并查集(map实现)、并查集(数组实现)的运行结果和运行时间");System.out.println("随机生成的二维矩阵规模 : " + row + " * " + col);start = System.currentTimeMillis();System.out.println("感染方法的运行结果: " + numIslands3(board1));end = System.currentTimeMillis();System.out.println("感染方法的运行时间: " + (end - start) + " ms");start = System.currentTimeMillis();System.out.println("并查集(map实现)的运行结果: " + numIslands1(board2));end = System.currentTimeMillis();System.out.println("并查集(map实现)的运行时间: " + (end - start) + " ms");start = System.currentTimeMillis();System.out.println("并查集(数组实现)的运行结果: " + numIslands2(board3));end = System.currentTimeMillis();System.out.println("并查集(数组实现)的运行时间: " + (end - start) + " ms");System.out.println();row = 10000;col = 10000;board1 = generateRandomMatrix(row, col);board3 = copy(board1);System.out.println("感染方法、并查集(数组实现)的运行结果和运行时间");System.out.println("随机生成的二维矩阵规模 : " + row + " * " + col);start = System.currentTimeMillis();System.out.println("感染方法的运行结果: " + numIslands3(board1));end = System.currentTimeMillis();System.out.println("感染方法的运行时间: " + (end - start) + " ms");start = System.currentTimeMillis();System.out.println("并查集(数组实现)的运行结果: " + numIslands2(board3));end = System.currentTimeMillis();System.out.println("并查集(数组实现)的运行时间: " + (end - start) + " ms");}}

拓展：如果matrix极大 设计一种可行的并行方案

空降问题 

 

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;// 本题为leetcode原题
// 测试链接：https://leetcode.com/problems/number-of-islands-ii/
// 所有方法都可以直接通过
public class Code03_NumberOfIslandsII {public static List<Integer> numIslands21(int m, int n, int[][] positions) {UnionFind1 uf = new UnionFind1(m, n);List<Integer> ans = new ArrayList<>();for (int[] position : positions) {ans.add(uf.connect(position[0], position[1]));}return ans;}public static class UnionFind1 {private int[] parent;private int[] size;//区别，变化之后不在抹去size，因为要用size做标记//size=0代表没有初始化过，没有空降过1private int[] help;private final int row;private final int col;private int sets;public UnionFind1(int m, int n) {row = m;col = n;sets = 0;int len = row * col;parent = new int[len];size = new int[len];help = new int[len];}private int index(int r, int c) {return r * col + c;}private int find(int i) {int hi = 0;while (i != parent[i]) {help[hi++] = i;i = parent[i];}for (hi--; hi >= 0; hi--) {parent[help[hi]] = i;}return i;}private void union(int r1, int c1, int r2, int c2) {if (r1 < 0 || r1 == row || r2 < 0 || r2 == row || c1 < 0 || c1 == col || c2 < 0 || c2 == col) {return;//检查越界}int i1 = index(r1, c1);int i2 = index(r2, c2);if (size[i1] == 0 || size[i2] == 0) {return;}int f1 = find(i1);int f2 = find(i2);if (f1 != f2) {if (size[f1] >= size[f2]) {size[f1] += size[f2];parent[f2] = f1;} else {size[f2] += size[f1];parent[f1] = f2;}sets--; }public int connect(int r, int c) {int index = index(r, c);if (size[index] == 0) {//为零代表第一次来到这个位置 parent[index] = index;size[index] = 1;sets++;union(r - 1, c, r, c);union(r + 1, c, r, c);union(r, c - 1, r, c);union(r, c + 1, r, c);}return sets;}}// 课上讲的如果m*n比较大，会经历很重的初始化，而k比较小，怎么优化的方法public static List<Integer> numIslands22(int m, int n, int[][] positions) {UnionFind2 uf = new UnionFind2();List<Integer> ans = new ArrayList<>();for (int[] position : positions) {ans.add(uf.connect(position[0], position[1]));}return ans;}public static class UnionFind2 {//m*n很大，key相对不大private HashMap<String, String> parent;//字符串他爹private HashMap<String, Integer> size;private ArrayList<String> help;private int sets;public UnionFind2() {parent = new HashMap<>();size = new HashMap<>();help = new ArrayList<>();sets = 0;}private String find(String cur) {while (!cur.equals(parent.get(cur))) {help.add(cur);cur = parent.get(cur);}for (String str : help) {parent.put(str, cur);}help.clear();return cur;}private void union(String s1, String s2) {if (parent.containsKey(s1) && parent.containsKey(s2)) {String f1 = find(s1);String f2 = find(s2);if (!f1.equals(f2)) {int size1 = size.get(f1);int size2 = size.get(f2);String big = size1 >= size2 ? f1 : f2;String small = big == f1 ? f2 : f1;parent.put(small, big);size.put(big, size1 + size2);sets--;}}}public int connect(int r, int c) {String key = String.valueOf(r) + "_" + String.valueOf(c);if (!parent.containsKey(key)) {//之前没空降过的parent.put(key, key);size.put(key, 1);sets++;String up = String.valueOf(r - 1) + "_" + String.valueOf(c);String down = String.valueOf(r + 1) + "_" + String.valueOf(c);String left = String.valueOf(r) + "_" + String.valueOf(c - 1);String right = String.valueOf(r) + "_" + String.valueOf(c + 1);union(up, key);union(down, key);union(left, key);union(right, key);}return sets;}}}