LeetCode-127. Word Ladder [C++][Java]

LeetCode-127. Word Ladderhttps://leetcode.com/problems/word-ladder/

题目描述

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

• Every adjacent pair of words differs by a single letter.
• Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
• sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

解题思路

BFS

【C++解法】

1. map + queue

此处仅限于计数

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector& wordList) {
        unordered_map> mp;
        for(string& w:wordList) for(int i=0; i q;
        q.push(beginWord);
        while(!q.empty()){
            ans++;
            queue nq;
            while(!q.empty()){
                string s = q.front(); q.pop();
                for(int i=0; i

2. three set

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector& wordList) {
        unordered_set dict(wordList.begin(), wordList.end());
        if(!dict.count(endWord)) return 0;
        unordered_set forward;
        forward.insert(beginWord);
        unordered_set backward;
        backward.insert(endWord);
        unordered_set next;
        int steps = 0;
        while (!forward.empty() && !backward.empty()) {
          ++steps;
          if (forward.size() > backward.size()) swap(forward, backward);
          for (auto &word : forward) dict.erase(word);
          for (auto &word: backward) dict.erase(word);
          for (string word : forward) {
            for (int i = 0; i < word.length(); ++i) {
              const char c = word[i];
              for (int j = 'a'; j <= 'z'; ++j) {
                if (c == j) continue;
                word[i] = j;
                if (backward.count(word)) return steps + 1; //恰好相邻
                if (!dict.count(word)) continue; //都不存在
                next.insert(word);
              }
              word[i] = c;
            }
          }
          swap(forward, next);
          next.clear();
        }
        return 0;
    }
};

3. set + queue

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector& wordList) {
        unordered_set dict(wordList.begin(),wordList.end());
        if (dict.find(endWord) == dict.end()) {return 0;}
        queue q; 
        q.push(beginWord);
		int len = 1; 
        while (!q.empty()){
            int size = q.size();
            while (size--){
                string last = q.front(); q.pop();
                if (last == endWord) {return len;}
                dict.erase(last);
                for (int i=0; i

【Java解法】

class Solution {
    public int ladderLength(String beginWord, String endWord, List wordList) {
        Set dict = new LinkedHashSet<>(wordList);
        if (!dict.contains(endWord)) {return 0;}
        Queue q = new LinkedList();
        q.offer(beginWord);
        int len = 1;
        while (!q.isEmpty()){
            int sz = q.size();
            while (sz-- > 0){
                String last = q.poll();
                if (last.equals(endWord)) {return len;}
                dict.remove(last);
                for (int i=0; i

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