LeetCode——Word Ladder II

链接：http://leetcode.com/onlinejudge#question_126

原题：

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.

思路：

其实解题思路和Word Ladder完全一样，BFS，但是麻烦的是要返回所有的路径。

所以没办法，只能把每个单词所对应的前驱单词记录下来，当然有可能有多个，那么

就用一个vector<string>存储好，有这些记录就可以重构路径了。

代码：

class Solution {public:    vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        pathes.clear();        dict.insert(start);dict.insert(end);vector<string> prev;unordered_map<string, vector<string> > traces;for (unordered_set<string>::const_iterator citr = dict.begin(); citr != dict.end(); citr++) {traces[*citr] = prev;}vector<unordered_set<string> > layers(2);int cur = 0;int pre = 1;layers[cur].insert(start);while (true) {cur = !cur;pre = !pre;for (unordered_set<string>::const_iterator citr = layers[pre].begin();citr != layers[pre].end(); citr++)dict.erase(*citr);layers[cur].clear();for (unordered_set<string>::const_iterator citr = layers[pre].begin();citr != layers[pre].end(); citr++) {for (int n=0; n<(*citr).size(); n++) {                      string word = *citr;                      int stop = word[n] - 'a';                      for (int i=(stop+1)%26; i!=stop; i=(i+1)%26) {                          word[n] = 'a' + i;                          if (dict.find(word) != dict.end()) {                              traces[word].push_back(*citr);            layers[cur].insert(word);                         }                      }                }}            if (layers[cur].size() == 0)                return pathes;if (layers[cur].count(end))break;}vector<string> path;buildPath(traces, path, end);return pathes;}private:void buildPath(unordered_map<string, vector<string> > &traces, vector<string> &path, const string &word) {if (traces[word].size() == 0) {                path.push_back(word);vector<string> curPath = path;reverse(curPath.begin(), curPath.end());pathes.push_back(curPath);                path.pop_back();return;}const vector<string> &prevs = traces[word];path.push_back(word);for (vector<string>::const_iterator citr = prevs.begin();citr != prevs.end(); citr++) {buildPath(traces, path, *citr);}path.pop_back();}vector<vector<string> > pathes;};