Leetcode 之图的算法

1. Clone Graph

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         / \         \_/

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。

UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        if(node == NULL) return NULL;    unordered_map<UndirectedGraphNode *, UndirectedGraphNode *>nodeMap;    queue<UndirectedGraphNode *> visit;    visit.push(node);    UndirectedGraphNode * nodeCopy = new UndirectedGraphNode(node->label);    nodeMap[node] = nodeCopy;    while(visit.size()>0){        UndirectedGraphNode * cur = visit.front();        visit.pop();        for(int i=0; i<cur->neighbors.size();i++){            UndirectedGraphNode *neighb = cur->neighbors[i];            if(nodeMap.find(neighb) == nodeMap.end()){                UndirectedGraphNode *neighCopy = new UndirectedGraphNode(neighb->label);                nodeMap[cur]->neighbors.push_back(neighCopy);                nodeMap[neighb] = neighCopy;                visit.push(neighb);            }else{                nodeMap[cur]->neighbors.push_back(nodeMap[neighb]);            }        }    }    return nodeCopy;    }