2014-04-25 20:18
题目：给定一个Node结构体，其中包含数据成员和两个Node*指针指向其他两个Node结构(还不如直接说这是个图呢)。给你一个Node指针作为参数，请做一份深拷贝作为结果返回。
解法：BFS搞定，需要检测重复节点以防止死循环，用一个哈希表可以做大。这样肯定只能找出一个完整的连通分量，其他连通分量的节点是无法检测到的。下面的代码其实是我在做leetcode时写的。
代码：
1 // 13.7 Given a pointer to a Node strcut, return a deep copy of whatever you can find with it. 2 // Answer: 3 // The following code is actually my solution to the leetcode problem, Clone Graph. 4 // They are different problems, but almost the same idea of BFS, mapping and deep copy. 5 #include <queue> 6 #include <vector> 7 #include <unordered_map> 8 using namespace std; 9 /** 10 * Definition for undirected graph. 11 * struct UndirectedGraphNode { 12 * int label; 13 * vector<UndirectedGraphNode *> neighbors; 14 * UndirectedGraphNode(int x) : label(x) {}; 15 * }; 16 */ 17 class Solution { 18 public: 19 UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) { 20 if (node == nullptr) { 21 return nullptr; 22 } 23 24 UndirectedGraphNode *cur, *nei; 25 int i, j; 26 int nc; 27 int ix, iy; 28 int nsize; 29 30 nc = 0; 31 qq.push(node); 32 while (!qq.empty()) { 33 cur = qq.front(); 34 qq.pop(); 35 if (um.find(cur) == um.end()) { 36 um[cur] = nc++; 37 graph.push_back(vector<int>()); 38 labels.push_back(cur->label); 39 nodes_checked.push_back(false); 40 } 41 ix = um[cur]; 42 if (nodes_checked[ix]) { 43 continue; 44 } 45 nsize = (int)cur->neighbors.size(); 46 for (i = 0; i < nsize; ++i) { 47 nei = cur->neighbors[i]; 48 if (um.find(nei) == um.end()) { 49 um[nei] = nc++; 50 labels.push_back(nei->label); 51 graph.push_back(vector<int>()); 52 nodes_checked.push_back(false); 53 } 54 iy = um[nei]; 55 if (!nodes_checked[iy]) { 56 qq.push(nei); 57 } 58 graph[ix].push_back(iy); 59 } 60 nodes_checked[ix] = true; 61 } 62 63 new_nodes.clear(); 64 for (i = 0; i < nc; ++i) { 65 new_nodes.push_back(new UndirectedGraphNode(labels[i])); 66 } 67 for (i = 0; i < nc; ++i) { 68 nsize = (int)graph[i].size(); 69 for (j = 0; j < nsize; ++j) { 70 new_nodes[i]->neighbors.push_back(new_nodes[graph[i][j]]); 71 } 72 } 73 cur = new_nodes[0]; 74 while (!qq.empty()) { 75 qq.pop(); 76 } 77 um.clear(); 78 new_nodes.clear(); 79 for (i = 0; i < (int)graph.size(); ++i) { 80 graph[i].clear(); 81 } 82 graph.clear(); 83 labels.clear(); 84 nodes_checked.clear(); 85 86 return cur; 87 } 88 private: 89 queue<UndirectedGraphNode *> qq; 90 unordered_map<UndirectedGraphNode *, int> um; 91 vector<int> labels; 92 vector<bool> nodes_checked; 93 vector<UndirectedGraphNode *> new_nodes; 94 vector<vector<int> > graph; 95 };