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 intermediate word must exist in the word list
For example,
Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.
从start变换到end，途中只能经过字典中的单词，每次只允许差一个字母。
要求输出所有最短变换路径。
这里刚开始使用了最普通的方法，即利用DFS，超时了。
public class Solution { List list = new ArrayList<List>(); int num = 0; String[] words; String[] ans; public List<List<String>> findLadders(String beginWord, String endWord, Set<String> wordList) { if( beginWord.length() == 0 || wordList.size() == 0 ) return list; int len = wordList.size(); words = new String[len]; int i = 0; for( String str : wordList ){ words[i] = str; i++; } ans = new String[len+2]; ans[0] = beginWord; helper(beginWord,endWord,1); return list; } public void helper(String beginWord,String endWord,int pos){ if( OneDiff(beginWord,endWord)){ if( num == 0 ) num = pos; else if( num < pos ) return; else if( num > pos ){ list.clear(); num = pos; } List ll = new ArrayList<String>(); for( int i = 0;i<pos;i++) ll.add(ans[i]); ll.add(endWord); list.add(ll); return ; } for( int i = 0;i<words.length;i++){ String str = words[i]; if( str == "-1") continue; if( OneDiff(beginWord,str) ){ ans[pos] = str; words[i] = "-1"; helper(str,endWord,pos+1); words[i] = str; } } } public boolean OneDiff(String word1,String word2){ int flag = 0; for( int i = 0;i<word1.length();i++){ if( word1.charAt(i) == word2.charAt(i) ) continue; else{ if( flag == 0) flag =1; else return false; } } return true; } } }
2、使用BFS，仍旧超时了。
public class Solution { List list = new ArrayList<List>(); Queue queue; HashMap map = new HashMap<String,Integer>(); public List<List<String>> findLadders(String beginWord, String endWord, Set<String> wordList) { if( beginWord.length() == 0 || wordList.size() == 0 ) return list; queue = new LinkedList<ArrayList<String>>(); if( OneDiff(beginWord,endWord) ){ List ll = new ArrayList<String>(); ll.add(beginWord); ll.add(endWord); list.add(ll); return list; } List ll = new ArrayList<String>(); ll.add(beginWord); int flag = 0; int deep = 0; map.put(beginWord,deep); queue.add(ll); while( !queue.isEmpty() ){ deep++; int len = queue.size(); for( int i = 0;i<len;i++) { List l1 = (List) queue.poll(); String s1 = (String) l1.get(l1.size() - 1); for (String str : wordList) { if (OneDiff(s1, str) && ( !map.containsKey(str) || (int)map.get(str) == deep) ) { map.put(str,deep); List l2 = new ArrayList<String>(); l2.addAll(l1); l2.add(str); queue.add(l2); if (OneDiff(str, endWord)) { l2.add(endWord); // for (int ii = 0; ii < l2.size(); ii++) // System.out.print(l2.get(ii) + " "); // System.out.println(); list.add(l2); flag = 1; } } } } if( flag == 1) queue.clear(); } return list; } public boolean OneDiff(String word1,String word2){ int flag = 0; for( int i = 0;i<word1.length();i++){ if( word1.charAt(i) == word2.charAt(i) ) continue; else{ if( flag == 0) flag++; else return false; } } if( flag == 1) return true; else return false; } }
3、借鉴了他人的答案
优化方法：先BFS生成找到end时的生成树，标记出每个单词所在的层数。然后从目标用DFS往回找，过了大数据。
NOTE：我在代码中刚开始由于存在static，导致测试用例一直无法通过。
public class Solution { List list = new ArrayList<List>(); HashMap map = new HashMap<String,Integer>(); public List<List<String>> findLadders(String beginWord, String endWord, Set<String> wordList) { if( beginWord == null || beginWord.length() == 0 || wordList.size() == 0 || beginWord.length() != endWord.length() ) return list; BFS(beginWord,endWord,wordList); if( map.size()== 1 ) return list; DFS(endWord,beginWord,new ArrayList<String>()); return list; } public void DFS(String beginWord,String endWord,List ans){ if( beginWord.equals(endWord) ){ ans.add(beginWord); Collections.reverse(ans); list.add(ans); return ; } if( map.get(beginWord) == null ) return ; ans.add(beginWord); int deep = (int) map.get(beginWord)-1; for( int i = 0;i<beginWord.length();i++){ char[] word = beginWord.toCharArray(); for( char ch = 'a';ch<='z';ch++) { word[i] = ch; String nWord = new String(word); if ( map.get(nWord) != null && ((int) map.get(nWord) == deep)) { ArrayList ll = new ArrayList<String>(ans); DFS(nWord, endWord, ll); } } } } public void BFS(String beginWord,String endWord,Set<String> wordList){ Queue queue = new LinkedList<String>(); queue.add(beginWord); map.put(beginWord,0); while( !queue.isEmpty() ){ String str = (String) queue.poll(); if( str.equals(endWord) ) continue; for( int i = 0 ;i <beginWord.length();i++){ char[] word = str.toCharArray(); for( char ch = 'a';ch<='z';ch++) { word[i] = ch; String Nword = new String(word); if ( Nword.equals(endWord) || wordList.contains(Nword)) { if (!map.containsKey(Nword)) { map.put(Nword, (int) map.get(str) + 1); queue.add(Nword); } } } } } } }