126. Word Ladder 2

当年这道题出来的时候是有名的难题。
我做的时候也是叹为观止。不过现在套路大家都熟了。
Word ladder就是图的问题，所以先建图
然后找所有的最短路径。
找最短路径的办法就是BFS。
这里由于我们要求出所有的最短路径所以要按level order来做。
我们做BFS的时候对于每个点它从哪里来的也记一下。
然后回溯回去重建路径就好了。
这么看的话也不是很难。
就是把大问题分解成小问题，然后分头解决小问题。
人的大脑容易虽然很大，可是注意力是有限的， 所以要分解成小问题
这就是一个BFS找最短路径然后重建所有最短路径的问题。

class Solution {
    public List> findLadders(String beginWord, String endWord, List wordList) {
        Map> graph = buildGraph(beginWord, wordList);

        //bfs to find shortest path and record upper streams
        Map> upperStreams = new HashMap<>();
        int dist = bfsHelper(beginWord, endWord, graph, upperStreams); 

        //reconstruct the path;
        List> ans = new ArrayList<>();
        if (dist == -1) return ans;

        LinkedList path = new LinkedList<>();
        path.add(endWord);
        dfsHelper(path, upperStreams, ans, beginWord); // reconstruct all paths
        return ans;
    }
    private void dfsHelper(LinkedList path, Map> upperStreams, 
                           List> ans, String beginWord ) {
        //base case;
        if (path.get(0).equals(beginWord)) {
            ans.add(new ArrayList<>(path));
            return;
        }
        String cur = path.get(0);
        for (String prev : upperStreams.get(cur)) {
            path.addFirst(prev);
            dfsHelper(path, upperStreams, ans, beginWord);
            path.removeFirst();
        }
    }
    
    private int bfsHelper(String beginWord, String endWord,  Map> graph,  Map> upperStreams ) {
        Queue queue = new LinkedList<>();
        queue.offer(beginWord);
        Set visited = new HashSet<>();
        visited.add(beginWord);
        int level = 1;
        boolean found = false;
        while (!queue.isEmpty()) {
            int size = queue.size();
            Set visitedAtThisLevel = new HashSet<>(); // need to collection all path so can't put to visited immediately
            for (int i = 0; i < size; i++) {
                String w = queue.poll();
                // expand;
                if (graph.get(w) == null) continue;
                for (String next : graph.get(w)) {
                    if (visited.contains(next)) continue;
                    upperStreams.putIfAbsent(next, new HashSet<>());
                    upperStreams.get(next).add(w);
                    if (next.equals(endWord)) found = true;
                    if (!visitedAtThisLevel.contains(next)) queue.add(next);
                    visitedAtThisLevel.add(next);
                }
            }
            visited.addAll(visitedAtThisLevel);
            if (found) return level;
            level++;
        }
        
        return -1;
    }
        
    private  Map> buildGraph(String beginWord, List wordList){
        wordList.add(beginWord);
        Map> graph = new HashMap<>();
        for (int i = 0; i < wordList.size(); i++) {
            for (int j = i + 1; j < wordList.size(); j++) {
                if (diffOne(wordList.get(i), wordList.get(j)) ) {
                    String w1 = wordList.get(i), w2 = wordList.get(j);
                    graph.putIfAbsent(w1, new HashSet<>());
                    graph.putIfAbsent(w2, new HashSet<>());
                    graph.get(w1).add(w2);
                    graph.get(w2).add(w1);
                }
            }
        }
        return graph;
    }
    private boolean diffOne(String w1, String w2) {
        int count = 0;
        for (int i = 0; i < w1.length(); i++) {
            if (w1.charAt(i) != w2.charAt(i)) count++;
            if (count > 1) return false;
        }
        return count == 1;
    }
}