LeetCode 127 Word Ladder (BFS)

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博客围绕给定两个单词和单词列表，求从起始单词到结束单词的最短转换序列长度展开。要求每次只能改变一个字母，转换后的单词需在列表中。将字符串哈希成数字并连边，转化为最短路问题求解，当前用时233ms，待优化。

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

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

Example 2:

Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

题目链接：https://leetcode.com/problems/word-ladder/

题目分析：常规操作，字符串hash成数字，根据题意连边，然后就是一个最短路问题

233ms，时间击败27.8%，待优化

class Solution {
    
    public final int MAX = 50000;
    
    public int[] head = new int[MAX];
    public int cnt = 0;
    HashMap<String, Integer> strHash = new HashMap<>();
    
    public boolean isLinked(String w1, String w2) {
        int eq = 1;
        for (int i = 0; i < w1.length(); i++) {
            if (w1.charAt(i) == w2.charAt(i)) {
                continue;
            }
            eq--;
            if (eq == -1) {
                return false;
            }
        }
        return true; 
    }
    
    public class Edge {
        public int to;
        public int nxt;
        
        public Edge(){}
        
        public Edge(int to, int nxt) {
            this.to = to;
            this.nxt = nxt;
        }
    }
    
    public Edge[] e = new Edge[MAX];
    
    public void Init() {
        cnt = 0;
        Arrays.fill(head, -1);
        for (int i = 0; i < MAX; i++) {
            e[i] = new Edge();
        }
    }
    
    public void Add(int u, int v) {
        e[cnt].to = v;
        e[cnt].nxt = head[u];
        head[u] = cnt++;
        
        e[cnt].to = u;
        e[cnt].nxt = head[v];
        head[v] = cnt++;
    }
    
    public int BFS(int st, int ed) {
        boolean vis[] = new boolean[MAX];
        Queue<int[]> q = new LinkedList();
        q.add(new int[]{st, 1});
        vis[st] = true;
        while (!q.isEmpty()) {
            int[] cur = q.poll();
            if (cur[0] == ed) {
                return cur[1];
            }
            for (int i = head[cur[0]]; i != -1; i = e[i].nxt) {
                int v = e[i].to;
                if (!vis[v]) {
                    vis[v] = true;
                    q.add(new int[]{v, cur[1] + 1});
                }
            }
        }

        return 0;
    }
    
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        int num = 1;
        strHash.put(beginWord, 0);
        for (int i = 0; i < wordList.size(); i++) {
            strHash.put(wordList.get(i), num++);
        }
        
        Init();
        
        for (int i = 0; i < wordList.size(); i++) {
            for (int j = i + 1; j < wordList.size(); j++) {
                if (isLinked(wordList.get(i), wordList.get(j))) {
                    int x = strHash.get(wordList.get(i));
                    int y = strHash.get(wordList.get(j));
                    Add(x, y);
                }
            }
        }
        for (int i = 0; i < wordList.size(); i++) {
            if (isLinked(beginWord, wordList.get(i))) {
                int x = strHash.get(wordList.get(i));
                Add(0, x);
            }
        }
        
        if (!strHash.containsKey(endWord)) {
            return 0;
        }

        return BFS(0, strHash.get(endWord));
    }
}