【leetcode】Word Ladder II
Word Ladder II
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
- Only one letter can be changed at a time
- Each intermediate word must exist in the dictionary
For example,
Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]
Return
[ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ]
Note:
- All words have the same length.
- All words contain only lowercase alphabetic characters.
先使用BFS,得到father,记录广度搜索路径上,每一个节点的父节点(可以有多个)
因此采用 unordered_map<string,unordered_set<string>> 数据结构
例如father["aaa"]=["baa","caa"]表示了aaa在广度优先搜素路径上的父节点为"baa","caa";
广度优先搜索时,从start开始,
找到与start相邻的节点 node1,node2,.....;
并记录每一个节点的父节点为start
然后从node1,node2...开始,遍历下一层
直到找到end节点停止(注意,必须上一层节点全部遍历完
找到father 路径后,从end开始往前dfs就可以得到所有的结果了
1 class Solution { 2 public: 3 vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) { 4 5 vector<vector<string>> result; 6 unordered_set<string> unvisited=dict; 7 8 dict.insert(start); 9 dict.insert(end); 10 11 unordered_map<string,unordered_set<string>> father; 12 if(unvisited.count(start)==0) unvisited.erase(start); 13 14 15 unordered_set<string> curString,nextString; 16 curString.insert(start); 17 18 19 while(curString.count(end)==0&&curString.size()>0) 20 { 21 22 for(auto it=curString.begin();it!=curString.end();it++) 23 { 24 string word=*it; 25 for(int i=0;i<word.length();i++) 26 { 27 string tmp=word; 28 for(int j='a';j<='z';j++) 29 { 30 if(tmp[i]==j) continue; 31 tmp[i]=j; 32 if(unvisited.count(tmp)>0) 33 { 34 nextString.insert(tmp); 35 father[tmp].insert(word); 36 37 } 38 39 } 40 } 41 } 42 43 if(nextString.size()==0) break; 44 45 for(auto it=nextString.begin();it!=nextString.end();it++) 46 { 47 //必须遍历完了curString中所有的元素,才能在unvisited中删除(因为可能有多个父节点对应着该节点) 48 unvisited.erase(*it); 49 } 50 51 curString=nextString; 52 nextString.clear(); 53 54 } 55 56 if(curString.count(end)>0) 57 { 58 vector<string> tmp; 59 dfs(father,end,start,result,tmp); 60 } 61 62 return result; 63 } 64 65 66 void dfs(unordered_map<string,unordered_set<string>> &father,string end,string start,vector<vector<string>> &result,vector<string> tmp) 67 { 68 tmp.push_back(end); 69 if(end==start) 70 { 71 reverse(tmp.begin(),tmp.end()); 72 result.push_back(tmp); 73 return; 74 } 75 76 for(auto it=father[end].begin();it!=father[end].end();it++) 77 { 78 dfs(father,*it,start,result,tmp); 79 } 80 } 81 };