Word Ladder II
<p>A <strong>transformation sequence</strong> from word <code>beginWord</code> to word <code>endWord</code> using a dictionary <code>wordList</code> is a sequence of words <code>beginWord -&gt; s<sub>1</sub> -&gt; s<sub>2</sub> -&gt; ... -&gt; s<sub>k</sub></code> such that:</p>

<ul>
	<li>Every adjacent pair of words differs by a single letter.</li>
	<li>Every <code>s<sub>i</sub></code> for <code>1 &lt;= i &lt;= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>
	<li><code>s<sub>k</sub> == endWord</code></li>
</ul>

<p>Given two words, <code>beginWord</code> and <code>endWord</code>, and a dictionary <code>wordList</code>, return <em>all the <strong>shortest transformation sequences</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]
<strong>Output:</strong> [[&quot;hit&quot;,&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;cog&quot;],[&quot;hit&quot;,&quot;hot&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]]
<strong>Explanation:</strong>&nbsp;There are 2 shortest transformation sequences:
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; &quot;cog&quot;
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;lot&quot; -&gt; &quot;log&quot; -&gt; &quot;cog&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;]
<strong>Output:</strong> []
<strong>Explanation:</strong> The endWord &quot;cog&quot; is not in wordList, therefore there is no valid transformation sequence.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= beginWord.length &lt;= 5</code></li>
	<li><code>endWord.length == beginWord.length</code></li>
	<li><code>1 &lt;= wordList.length &lt;= 500</code></li>
	<li><code>wordList[i].length == beginWord.length</code></li>
	<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>
	<li><code>beginWord != endWord</code></li>
	<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>
	<li>The <strong>sum</strong> of all shortest transformation sequences does not exceed <code>10<sup>5</sup></code>.</li>
</ul>
Breadth-First Search; Hash Table; String; Backtracking
func findLadders(beginWord string, endWord string, wordList []string) [][]string {
	var ans [][]string
	words := make(map[string]bool)
	for _, word := range wordList {
		words[word] = true
	}
	if !words[endWord] {
		return ans
	}
	words[beginWord] = false
	dist := map[string]int{beginWord: 0}
	prev := map[string]map[string]bool{}
	q := []string{beginWord}
	found := false
	step := 0
	for len(q) > 0 && !found {
		step++
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			chars := []byte(p)
			for j := 0; j < len(chars); j++ {
				ch := chars[j]
				for k := 'a'; k <= 'z'; k++ {
					chars[j] = byte(k)
					t := string(chars)
					if v, ok := dist[t]; ok {
						if v == step {
							prev[t][p] = true
						}
					}
					if !words[t] {
						continue
					}
					if len(prev[t]) == 0 {
						prev[t] = make(map[string]bool)
					}
					prev[t][p] = true
					words[t] = false
					q = append(q, t)
					dist[t] = step
					if endWord == t {
						found = true
					}
				}
				chars[j] = ch
			}
		}
	}
	var dfs func(path []string, begin, cur string)
	dfs = func(path []string, begin, cur string) {
		if cur == beginWord {
			cp := make([]string, len(path))
			copy(cp, path)
			ans = append(ans, cp)
			return
		}
		for k := range prev[cur] {
			path = append([]string{k}, path...)
			dfs(path, beginWord, k)
			path = path[1:]
		}
	}
	if found {
		path := []string{endWord}
		dfs(path, beginWord, endWord)
	}
	return ans
}