[Leetcode] 126. 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.

这道题感觉是LC中最难的题之一，先用BFS来构建undirected word graph，同时把每层的单词到start的距离也保存了。

第二步再用回溯+dfs的方法把一条条路径给找出来。

import java.util.ArrayList;import java.util.HashMap;import java.util.LinkedList;public class Solution {    public ArrayList<ArrayList<String>> findLadders(String start, String end, Set<String> dict) {        ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();        HashMap<String, ArrayList<String>> wordGraph = new HashMap<String, ArrayList<String>>();        HashMap<String, Integer> distance = new HashMap<String, Integer>();        dict.add(start);        dict.add(end);        bfs(start, end, dict, wordGraph, distance);        ArrayList<String> list = new ArrayList<String>();        list.add(start);        dfs(start, end, wordGraph, result, list, distance);        return result;    }    private void dfs(String current, String end, HashMap<String, ArrayList<String>> wordGraph,         ArrayList<ArrayList<String>> result, ArrayList<String> list, HashMap<String, Integer> distance){        if(current.equals(end)){            result.add(new ArrayList<String>(list));        } else {            for(String next: wordGraph.get(current)){                if(distance.get(current) == distance.get(next) - 1){                    list.add(next);                    dfs(next, end, wordGraph, result, list, distance);                    list.remove(list.size() - 1);                }            }        }    }    private void bfs(String start, String end, Set<String> dict, HashMap<String,        ArrayList<String>> wordGraph, HashMap<String, Integer> distance){        Queue<String> queue = new LinkedList<String>();        queue.offer(start);        distance.put(start, 0);        while(!queue.isEmpty()){            String current = queue.poll();            ArrayList<String> nextList = findAdjacent(current, dict);            wordGraph.put(current, nextList);            for(String next: nextList){                if(!distance.containsKey(next)){                    distance.put(next, distance.get(current) + 1);                    queue.offer(next);                }            }        }    }    private ArrayList<String> findAdjacent(String s, Set<String> dict){        ArrayList<String> result = new ArrayList<String>();        for(int i = 0; i < s.length(); i++){            for(char c = 'a'; c <= 'z'; c++){                if(c == s.charAt(i)){                    continue;                }                String temp = s.substring(0, i) + c + s.substring(i + 1);                if(dict.contains(temp)){                    result.add(temp);                }            }        }        return result;    }}