[LeetCode] Word Ladder

Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence 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"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.

LeetCode: https://oj.leetcode.com/problems/word-ladder/

這道題的求解思路爲：對於start，首先找出dict中和start只有一個字符不同的字符串集合S1，對於該S1中的字符串，start到他們的最短距離爲1；然後再求解dict中和S1中字符串只相差一個字符並且沒有出現在S1中的字符集合S2；則start到S2中的字符的最短距離爲2，通過上述過程計算循環計算下去，直到某個集合Si中出現end，則找到start到end的最短距離。

其實敏感的同學可以發現者其實是一顆以start爲根的樹，樹中的節點包括start、end、dict中的字符串（如果包含end，則end一定爲葉節點，但有可能出現在多個位置上）；本題的目標其實是要找到從start到end的最短路徑；對於這類問題，一般都採用DFS或BFS，由於樹本身並沒有事先建立好，所以不太適合採用DFS；而通過上面求解思路可知，採用BFS比較恰當。

代碼如下：

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        if(start == null || end == null || start.length() == 0 || end.length() == 0 || start.length() != end.length())
            return 0;
        Queue<String> queue = new LinkedList<String>();
        HashSet<String> visited = new HashSet<String>();
        queue.offer(start);
        int depth = 1;
        int cur = 1;
        int next = 0;
        while(!queue.isEmpty()){
            String e = queue.poll();
            cur--;
            for(int i = 0; i < e.length(); i++){
                char[] ca = e.toCharArray();
                for(char c = 'a'; c <= 'z'; c++){
                    ca[i] = c;
                    String s = new String(ca);
                    if(s.equals(end))
                        return depth+1;
                    if(dict.contains(s) && !visited.contains(s)){
                        queue.offer(s);
                        visited.add(s);
                        next++;
                    }
                }
            }
            if(cur == 0){
                cur = next;
                next = 0;
                depth++;
            }
        }
        return 0;
    }
}