标签：seq ack back return hat amp set link and
Given two words (beginWord and endWord), and a dictionary‘s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: Only one letter can be changed at a time Each transformed word must exist in the word list. Note that beginWord is not a transformed word. For example, Given: beginWord = "hit" endWord = "cog" wordList = ["hot","dot","dog","lot","log","cog"] Return [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ] Note: Return an empty list if there is no such transformation sequence. All words have the same length. All words contain only lowercase alphabetic characters. You may assume no duplicates in the word list. You may assume beginWord and endWord are non-empty and are not the same.
linkedin : word ladder 1.5 如果被challenge用26个char iteration太慢的话，大家怎么optimize？
在word ladder 基础上 就用了parent hash map backtracking 可以吗？
public class WordLadder15 {
Map<String, Set<String>> map = new HashMap<>();
Map<String, String> pmap = new HashMap<>(); //parents
public List<String> ladderLength(String bw, String ew, List<String> words) {
List<String> list = new ArrayList<>(words);
if (bw.equals(ew))
return null;
list.add(bw);
list.add(ew);
// build graph. From 1point 3acres bbs
for (String w : list) {
Set<String> set = new HashSet<>();
for (String w2 : list) {
if (!w2.equals(w) && connect(w2, w))
set.add(w2);
}
map.put(w, set);
}
Queue<String> q = new ArrayDeque<>();
Set<String> seen = new HashSet<>();
q.offer(bw);
seen.add(bw);. From 1point 3acres bbs
while (!q.isEmpty()) {
int len = q.size();
for (int i = 0; i < len; i++) {
String cur = q.poll();
Set<String> set = map.get(cur);
for (String nei : set) {
if (seen.add(nei)){
pmap.put(nei, cur);
q.offer(nei);
}
if (nei.equals(ew))
return getPath(nei);
}
}
}
return new ArrayList<String>();
}
private List<String> getPath(String end){
List<String> res = new ArrayList<>();
String parent=end;
while(parent!=null){
res.add(0, parent);
parent=pmap.get(parent);
}. From 1point 3acres bbs
return res;
}
// build the graph: which connects with which
private boolean connect(String a, String b) {
int diff = 0;
for (int i = 0; i < a.length(); i++) {
if (a.charAt(i) != b.charAt(i))
diff++;
}
return diff == 1;
}
public static void main(String[] args) {
WordLadder15 sol = new WordLadder15();
List<String> res= sol.ladderLength("hit", "cog",
new ArrayList<String>(Arrays.asList("hot", "dot", "dog", "lot", "log", "cog")));
for(String s: res) System.out.println(" "+ s);
}
}
标签：seq ack back return hat amp set link and
原文地址：http://www.cnblogs.com/apanda009/p/7947633.html
