[Leetcode] Alien Dictionary, Solution

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:
Given the following words in dictionary,

[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

The correct order is: "wertf".

Example 2:
Given the following words in dictionary,

[
  "z",
  "x"
]

The correct order is: "zx".

Example 3:
Given the following words in dictionary,

[
  "z",
  "x",
  "z"
]

The order is invalid, so return "".

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

[Thoughts]
有向图的拓扑排序。进来的边为入度，出去的边为出度。生成图的结构，然后按照如下规则排序：
1. 找出当前入度为0的节点
2. 所有和该节点相连接的节点入度减1
3. go to #1

看下图示例：

[Code]

1:  string alienOrder(vector<string>& words) {  
2:       unordered_map<char, set<char>> outbound;  
3:       unordered_map<char, set<char>> inbound;  
4:    set<char> no_pre;  
5:      
6:    string s = "";  
7:    for(int i =0; i< words.size(); i++) {  
8:      string t = words[i];  
9:      no_pre.insert(t.begin(), t.end());  
10:      for(int j = 0; j< min(s.size(), t.size()); j++) {  
11:        if(t[j] != s[j]) {  
12:          inbound[t[j]].insert(s[j]);  
13:          outbound[s[j]].insert(t[j]);  
14:          break;  
15:        }  
16:      }  
17:      s = t;  
18:    }  
19:      
20:    // get the nodes which has 0 inbound edge  
21:    int char_count = no_pre.size();  
22:    for(auto p : inbound) {  
23:      no_pre.erase(p.first);  
24:    }  
25:      
26:    string result = "";  
27:    while(no_pre.size() >0) {  
28:      auto it = no_pre.begin();  
29:      char c = *it;  
30:      no_pre.erase(c);  
31:      result+=c;  
32:        
33:      for(char su : outbound[c]) {  
34:        inbound[su].erase(c);  
35:        if(inbound[su].size() == 0) {  
36:          no_pre.insert(su);  
37:        }  
38:      }  
39:        
40:    }   
41:       return result.size() == char_count? result : "";  
42:  }  

[LeetCode] Clone Graph, Solution

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

 

[Thoughts]

这题和链表拷贝类似：http://fisherlei.blogspot.com/2013/11/leetcode-copy-list-with-random-pointer.html

所不同的是，在链表拷贝中，没有借助额外空间，通过多次链表遍历来拷贝、链接及拆分。

而这里图的拷贝，也可以通过多次遍历来插入拷贝节点，链接拷贝节点以及将拷贝节点拆分出来。但是同样的问题是，需要对图进行多次遍历。如果想在一次遍历中，完成拷贝的话，那就需要使用额外的内存来使用map存储源节点和拷贝节点之间的对应关系。有了这个关系之后，在遍历图的过程中，就可以同时处理访问节点及访问节点的拷贝节点，一次完成。详细看下面代码。

 

[Code]

 1 /**
 2  * Definition for undirected graph.
 3  * struct UndirectedGraphNode {
 4  *     int label;
 5  *     vector<UndirectedGraphNode *> neighbors;
 6  *     UndirectedGraphNode(int x) : label(x) {};
 7  * };
 8  */
 9 class Solution {
10 public:
11     UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
12         if(node == NULL) return NULL;
13         unordered_map<UndirectedGraphNode *, UndirectedGraphNode *> nodeMap;
14         queue<UndirectedGraphNode *> visit;
15         visit.push(node);
16         UndirectedGraphNode * nodeCopy = new UndirectedGraphNode(node->label);
17         nodeMap[node] = nodeCopy;
18         while (visit.size()>0)
19         {
20             UndirectedGraphNode * cur = visit.front();
21             visit.pop();
22             for (int i = 0; i< cur->neighbors.size(); ++i)
23             {
24                 UndirectedGraphNode * neighb = cur->neighbors[i];
25                 if (nodeMap.find(neighb) == nodeMap.end())
26                 {
27                     // no copy of neighbor node yet. create one and associate with the copy of cur
28                     UndirectedGraphNode* neighbCopy = new UndirectedGraphNode(neighb->label);
29                     nodeMap[cur]->neighbors.push_back(neighbCopy);
30                     nodeMap[neighb] = neighbCopy;
31                     visit.push(neighb);
32                 }
33                 else
34                 {
35                     // already a copy there. Associate it with the copy of cur
36                     nodeMap[cur]->neighbors.push_back(nodeMap[neighb]);
37                 }
38             }
39         }
40     
41         return nodeCopy;
42     }
43 };