leetcode 87: Word Ladder II

Word Ladder II                                             Feb 11

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) fromstart toend, 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]

同word ladder I 一样, 但是为了能够trace back. 建立一个树, 树的节点有一个父指针. 同时, 删除字典中的word时, 不能像word ladder I中,立刻删, 因为同一个word有可能再本层中被别的branch使用, 所以, 统一在每层结束时删除. 用特殊字符"#", 标示每一层结束.

[CODE]

public class Solution {    public List<List<String>> findLadders(String start, String end, Set<String> dict) {        Queue<Node> que = new LinkedList<Node>();        List<List<String>> res = new ArrayList<List<String>>();        Set<String> rmList = new HashSet<String>();        dict.add(end);        boolean flag = false;        dict.remove(start);        que.add( new Node(start, null) );        que.add(new Node("#",null));                while(!que.isEmpty() ) {            Node n = que.remove();            if(n.val.equals("#") ) {                if(flag || que.isEmpty()) return res;                dict.removeAll(rmList);                rmList.clear();                que.add(new Node("#", null));                continue;            }                        if(n.val.equals(end) ) {                flag = true;                List<String> cur = getBranch(n);                res.add(cur);            }                        List<Node> adjs = findAdjs(n, end, dict, rmList);            que.addAll(adjs);        }        return res;    }        private List<String> getBranch(Node n) {        List<String> branch = new ArrayList<String>();        while(n!=null) {            branch.add(n.val);            n = n.parent;        }        Collections.reverse(branch);        return branch;    }        private List<Node> findAdjs(Node n, String end, Set<String> dict, Set<String> rmList) {    String s = n.val;        List<Node> adjs = new ArrayList<Node>();        for(int i=0; i<s.length(); i++) {            StringBuilder sb = new StringBuilder(s);            for(char c='a'; c<'z'; c++) {                sb.setCharAt(i, c);                String sbs = sb.toString();                if(dict.contains(sbs)) {                    adjs.add(new Node(sbs, n));                    if(!sbs.equals(end)) rmList.add(sbs);                }                 }        }        return adjs;    }}class Node{    String val;    Node parent;        public Node(String val, Node parent) {        this.val = val;        this.parent = parent;    }}