本文主要是介绍Topological sort题型总结,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
拓扑排序是对有向无环图的顶点的一种排序,
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检测编译时的循环依赖
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制定有依赖关系的任务的执行顺序
拓扑排序的算法是典型的宽度优先搜索算法,其大致流程如下:
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统计所有点的入度,并初始化拓扑序列为空。
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将所有入度为 0 的点,也就是那些没有任何依赖的点,放到宽度优先搜索的队列中
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将队列中的点一个一个的释放出来,放到拓扑序列中,每次释放出某个点 A 的时候,就访问 A 的相邻点(所有A指向的点),并把这些点的入度减去 1。
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如果发现某个点的入度被减去 1 之后变成了 0,则放入队列中。
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直到队列为空时,算法结束,
1. graph怎么构建
Graph 一般是adjecent list,
class DirectedGraphNode {
int label;
List<DirectedGraphNode> neighbors;
...
}
也可以使用 HashMap 和 HashSet 搭配的方式来存储邻接表
hashmap<Integer, List<Integer>>() 或者HashMap<Integer, HashSet<Integer>>() 来表示;
2.indegree怎么构建
array 或者hashmap<Node, Integer>
3. queue,构建拓扑排序,每次pop 入度为0的node;
首先来个经典的题目: Topological sort;
For graph as follow:
The topological order can be:
[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
思路:标准topo排序的算法;算indegree,然后每次remove node,neighbor的入度全部减1,以此循环;
/*** Definition for Directed graph.* class DirectedGraphNode {* int label;* ArrayList<DirectedGraphNode> neighbors;* DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }* };*/public class Solution {/** @param graph: A list of Directed graph node* @return: Any topological order for the given graph.*/public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {ArrayList<DirectedGraphNode> list = new ArrayList<DirectedGraphNode>();HashMap<DirectedGraphNode, Integer> indegree = new HashMap<DirectedGraphNode, Integer>();for(DirectedGraphNode node: graph) {// node -> neighbor, neighbor indegree + 1 ;for(DirectedGraphNode neighbor: node.neighbors) {if(indegree.containsKey(neighbor)){indegree.put(neighbor, indegree.get(neighbor) + 1);} else {indegree.put(neighbor, 1);}}}Queue<DirectedGraphNode> queue = new LinkedList<DirectedGraphNode>();for(DirectedGraphNode node: graph) {// if indegree not contains, means indegree is 0;if(!indegree.containsKey(node)){queue.offer(node);}}while(!queue.isEmpty()) {DirectedGraphNode node = queue.poll();list.add(node);for(DirectedGraphNode neighbor : node.neighbors) {indegree.put(neighbor, indegree.get(neighbor) - 1);if(indegree.get(neighbor) == 0) {queue.offer(neighbor);}}}return list;}
}
Course Schedule
[1,0] 代表的物理意义是:0 --> 1
思路:用hashmap build adjecent list,然后记录indegree,neighbor indegree相互减1. return count == numCourses.
class Solution {public boolean canFinish(int numCourses, int[][] prerequisites) {if(prerequisites == null || numCourses < 0) {return false;}int[] indegree = new int[numCourses];HashMap<Integer, List<Integer>> graph = new HashMap<>();for(int i = 0; i < numCourses; i++) {graph.putIfAbsent(i, new ArrayList<>());}for(int[] prerequist : prerequisites) {int course = prerequist[0];int pcourse = prerequist[1];// pcourse -> course;graph.get(pcourse).add(course);indegree[course]++;}Queue<Integer> queue = new LinkedList<>();for(int i = 0; i < numCourses; i++) {if(indegree[i] == 0) {queue.offer(i);}}int count = 0;while(!queue.isEmpty()) {Integer node = queue.poll();count++;for(Integer neighbor: graph.get(node)) {indegree[neighbor]--;if(indegree[neighbor] == 0) {queue.offer(neighbor);}}}return count == numCourses;}
}
Alien Dictionary 思路:从单词之间的关系来得到图的关系, 注意所有的char都是一个node,都是字母;然后用hashmap<Character, HashSet<Character>>来建立图。注意分函数写程序,这样清晰;注意indegree需要把每个node全部赋值为0;然后再进行+1;
class Solution {public String alienOrder(String[] words) {if(words == null || words.length == 0) {return "";}HashMap<Character, HashSet<Character>> graph = new HashMap<>();// 每个character,都得在graph里面建立一个node;for(String word: words) {for(int i = 0; i < word.length(); i++) {graph.putIfAbsent(word.charAt(i), new HashSet<Character>());}}for(int i = 0; i < words.length - 1; i++) {String worda = words[i];String wordb = words[i + 1];int minlen = Math.min(worda.length(), wordb.length());for(int j = 0; j < minlen; j++) {char ac = worda.charAt(j);char bc = wordb.charAt(j);graph.putIfAbsent(ac, new HashSet<Character>());graph.putIfAbsent(bc, new HashSet<Character>());// ac -> bc;if(ac != bc) {graph.get(ac).add(bc);break;}}// 特殊情况return abc -> ab, return "";if(worda.length() > wordb.length() && worda.substring(0, minlen).equals(wordb.substring(0, minlen))) {return "";}}HashMap<Character, Integer> indegree = new HashMap<>();buildIndegree(indegree, graph);Queue<Character> queue = new LinkedList<Character>();StringBuilder sb = new StringBuilder();for(Character node: indegree.keySet()) {if(indegree.get(node) == 0) {queue.offer(node);}}while(!queue.isEmpty()) {Character node = queue.poll();sb.append(node);for(Character neighbor: graph.get(node)) {indegree.put(neighbor, indegree.get(neighbor) - 1);if(indegree.get(neighbor) == 0) {queue.offer(neighbor);}}}return sb.length() == indegree.keySet().size() ? sb.toString() : "";}private void buildIndegree(HashMap<Character, Integer> indegree,HashMap<Character, HashSet<Character>> graph) {// graph的每个node,indegree初始为0;for(Character node: graph.keySet()) {indegree.putIfAbsent(node, 0);}for(Character node: graph.keySet()) {for(Character neighbor: graph.get(node)) {indegree.put(neighbor, indegree.getOrDefault(neighbor, 0) + 1);}}}
}
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