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图与搜索

时间:2016-10-23 14:59:05      阅读:275      评论:0      收藏:0      [点我收藏+]

标签:区别   bool   ons   简单图   ops   unique   .com   sort   else   

主要知识点:

克隆图

拓扑排序

DFS BFS

BFS两个使用场景:图的遍历 简单图求最短路径

BFS in Graph 和BFS in Tree的主要区别就是有无环

 

Clone Graph --not bug free

方法一:递归

技术分享
 1 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 2         // write your code here
 3         if (node == null) {
 4             return node;
 5         }
 6         HashMap<UndirectedGraphNode, UndirectedGraphNode> hashMap = new HashMap<>();
 7         return help(node, hashMap);
 8     }
 9     
10     public UndirectedGraphNode help(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> hashMap) {
11         UndirectedGraphNode root = new UndirectedGraphNode(node.label);
12         hashMap.put(node, root);
13         for (UndirectedGraphNode neighbor : node.neighbors) {
14             if (hashMap.containsKey(neighbor)) {
15                 root.neighbors.add(hashMap.get(neighbor));
16             } else {
17                 root.neighbors.add(help(neighbor, hashMap));
18             }
19         }
20         return root;
21     }
View Code

方法二:非递归,分两步:找节点和mapping  neighbors

技术分享
 1 public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
 2         // write your code here
 3         if (node == null) {
 4             return node;
 5         }
 6         //getNodeAndMapping
 7         List<UndirectedGraphNode> nodes = new ArrayList<UndirectedGraphNode>();
 8         HashMap<UndirectedGraphNode, UndirectedGraphNode> mapping = new HashMap<>();
 9         Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
10         queue.offer(node);
11         nodes.add(node);
12         mapping.put(node, new UndirectedGraphNode(node.label));
13         while (!queue.isEmpty()) {
14             UndirectedGraphNode cur = queue.poll();
15             for (UndirectedGraphNode neighbor : cur.neighbors) {
16                 if (mapping.containsKey(neighbor)) {
17                     continue;
18                 }
19                 nodes.add(neighbor);
20                 mapping.put(neighbor, new UndirectedGraphNode(neighbor.label));
21                 queue.offer(neighbor);
22             }
23         }
24         
25         //clone neighbor
26         for (UndirectedGraphNode old : nodes) {
27             UndirectedGraphNode newNode = mapping.get(old);
28             for (UndirectedGraphNode neighbor : old.neighbors) {
29                 newNode.neighbors.add(mapping.get(neighbor));
30             }
31         }
32         return mapping.get(node);
33     }
View Code

 

Topological Sorting

技术分享
 1 public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
 2         // write your code here
 3         Map<DirectedGraphNode, Integer> indegreeMap = new HashMap<DirectedGraphNode, Integer>();
 4         getInDegree(graph, indegreeMap);
 5         Queue<DirectedGraphNode> queue = new LinkedList<DirectedGraphNode>();
 6         for (DirectedGraphNode node : graph) {
 7             if (!indegreeMap.containsKey(node)) {
 8                 queue.offer(node);
 9             }
10         }
11         ArrayList<DirectedGraphNode> result = new ArrayList<>();
12         while (!queue.isEmpty()) {
13             DirectedGraphNode cur = queue.poll();
14             result.add(cur);
15             for (DirectedGraphNode neighbor : cur.neighbors) {
16                 int indegree = indegreeMap.get(neighbor);
17                 indegreeMap.put(neighbor, indegree - 1);
18                 if (indegree == 1) {
19                     queue.offer(neighbor);
20                 }
21             }
22         }
23         return result;
24     }
25     
26     public void getInDegree(ArrayList<DirectedGraphNode> graph,  
27         Map<DirectedGraphNode, Integer> indegreeMap) {
28             for (DirectedGraphNode node : graph) {
29                 for (DirectedGraphNode neighbor : node.neighbors) {
30                     if (!indegreeMap.containsKey(neighbor)) {
31                         indegreeMap.put(neighbor, 1);
32                     } else {
33                         indegreeMap.put(neighbor, indegreeMap.get(neighbor) + 1);
34                     }
35                 }
36             }   
37     }
View Code

 

Permutations

Given a list of numbers, return all possible permutations.You can assume that there is no duplicate numbers in the list.

技术分享
 1  public List<List<Integer>> permute(int[] nums) {
 2         // write your code here
 3         List<List<Integer>> results = new ArrayList<>();
 4         if (nums == null || nums.length == 0) {
 5             results.add(new ArrayList<Integer>());
 6             return results;
 7         }
 8         boolean[] used = new boolean[nums.length];
 9         BFS(nums, results, new ArrayList<Integer>(), used);
10         return results;
11     }
12     
13     public void BFS(int[] nums, List<List<Integer>> results, List<Integer> cur, boolean[] used) {
14         if (cur.size() == nums.length) {
15             results.add(new ArrayList<>(cur));
16             return ;
17         }
18         for (int i = 0; i < nums.length; i++) {
19             if (used[i]) {
20                 continue;
21             }
22             used[i] = true;
23             cur.add(nums[i]);
24             BFS(nums, results, cur, used);
25             cur.remove(cur.size() - 1);
26             used[i] = false;
27         }
28     }
View Code

 

Permutations II

注意不要忘了排序。

Given a list of numbers with duplicate number in it. Find all unique permutations.

技术分享
 1 public List<List<Integer>> permuteUnique(int[] nums) {
 2         // write your code here
 3         List<List<Integer>> results = new ArrayList<>();
 4         if (nums == null || nums.length == 0) {
 5             results.add(new ArrayList<Integer>());
 6             return results;
 7         }
 8         Arrays.sort(nums);//Attention!
 9         boolean[] used = new boolean[nums.length];
10         BFS(nums, results, new ArrayList<Integer>(), used);
11         return results;
12     }
13     
14     public void BFS(int[] nums, List<List<Integer>> results, List<Integer> cur, boolean[] used) {
15         if (cur.size() == nums.length) {
16             results.add(new ArrayList<>(cur));
17             return ;
18         }
19         for (int i = 0; i < nums.length; i++) {
20             if (used[i]) {
21                 continue;
22             }
23             if (i > 0 && !used[i - 1] && nums[i] == nums[i - 1]) {
24                 continue;
25             }
26             used[i] = true;
27             cur.add(nums[i]);
28             BFS(nums, results, cur, used);
29             cur.remove(cur.size() - 1);
30             used[i] = false;
31         }
32     }
View Code

 

图与搜索

标签:区别   bool   ons   简单图   ops   unique   .com   sort   else   

原文地址:http://www.cnblogs.com/futurehau/p/5989835.html

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