LeetCode OJ Word Ladder II

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

1. Only one letter can be changed at a time
2. 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.

看了Discuss里面的题解，https://leetcode.com/discuss/9523/share-two-similar-java-solution-that-accpted-by-oj

自己改成了C++版本，具体的思路就是把每个string当做一个点，先BFS找到边，然后DFS找路径，感觉有点像某道什么素数路径的题。

class Solution {public:vector<vector<string> > ans;  // answervector<string> backtraceList;  //  list for backtracing, maybe an answermap<string, vector<string> > edges;  // treat every string as a pointstring st;  // start stringstring ed;  // end stringunordered_set<string> visit;  // BFS visitunordered_set<string> unvisit;  // BFS unvisitvector<vector<string>> findLadders(string start, string end, unordered_set<string> & dict) {ans.clear();backtraceList.clear();edges.clear();if (dict.size() == 0) return ans;st = start;ed = end;visit.clear();unvisit = dict;BFS();  // bfs makes graphDFS(ed);  // dfs find roads(from end to start)return ans;}void BFS() {int cur = 1;  // current points to checkint next = 0;  // next points to checkint l;  // the length of each wordbool found = false;queue<string> q;q.push(st);unvisit.insert(ed);unvisit.erase(st);  // maybe there is another start string in the dictwhile (!q.empty()) {string w = q.front();q.pop();cur--;l = w.size();for (int i = 0; i < l; i++) {char originalChar = w[i];string nw = w;for (char c = 'a'; c <= 'z'; c++) {nw[i] = c;if (unvisit.find(nw) != unvisit.end()) {  // an unvisited wordif (visit.find(nw) == visit.end()) {  // there is no this word in visitvisit.insert(nw);next++;  // more word to visit next timeq.push(nw);}map<string, vector<string> >::iterator iter = edges.find(nw);if (iter != edges.end()) {  // a new point of an existed edge(*iter).second.push_back(w);}else {  // a new edgevector<string> temp;temp.push_back(w);edges[nw] = temp;}if (nw == ed) found = true;}}w[i] = originalChar;}if (cur == 0) {if (found) break;cur = next;next = 0;unordered_set<string>::iterator iter = visit.begin();for (; iter != visit.end(); iter++) unvisit.erase(*iter);visit.clear();}}}void DFS(string w) {if (w == st) {  // find start stringvector<string> temp;temp.push_back(st);for (int i = 0, s = backtraceList.size(); i < s; i++)temp.push_back(backtraceList[s - i - 1]);ans.push_back(temp);return;}backtraceList.push_back(w);map<string, vector<string> >::iterator iter = edges.find(w);if (iter != edges.end()) {int s = iter->second.size();for (int i = 0; i < s; i++) DFS(iter->second[i]);}backtraceList.pop_back();}};