[LeetCode] 127. 单词接龙
题目链接 : https://leetcode-cn.com/problems/word-ladder/
题目描述:
给定两个单词（beginWord 和 endWord）和一个字典，找到从 beginWord 到 endWord 的最短转换序列的长度。转换需遵循如下规则：
- 每次转换只能改变一个字母。
- 转换过程中的中间单词必须是字典中的单词。
说明:
- 如果不存在这样的转换序列，返回 0。
- 所有单词具有相同的长度。
- 所有单词只由小写字母组成。
- 字典中不存在重复的单词。
- 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。
示例:
示例 1:
输入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
输出: 5
解释: 一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog",
返回它的长度 5。
示例 2:
输入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
输出: 0
解释: endWord "cog" 不在字典中，所以无法进行转换。
思路:
一看这一题就是BFS,
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
from collections import deque
wordList = set(wordList)
if endWord not in wordList:
return 0
visited = set()
# 看两个单词是否差一个字母
def is_one_letter(w, word):
if len(w) != len(word) :
return False
dif = 0
for i in range(len(w)):
if w[i] != word[i]:dif += 1
if dif > 1: return False
return True
# 发现所有和word相差一个字母的单词
def find_only_one_letter(word):
ans = []
for w in wordList:
#print(is_one_letter(w, word))
if w not in visited and is_one_letter(w, word):
ans.append(w)
#print(ans)
return ans
queue = deque()
queue.appendleft(beginWord)
res = 1
# BFS
while queue:
n = len(queue)
for _ in range(n):
tmp = queue.pop()
if tmp == endWord:
return res
visited.add(tmp)
for t in find_only_one_letter(tmp):
queue.appendleft(t)
res += 1
return 0
但是, 上面才过了29/40测试用例,可能是由于每次都要和wordList比较是否是差一个单词, 换个其他方法.
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
from collections import deque
wordList = set(wordList)
if endWord not in wordList:
return 0
visited = set()
queue = deque()
queue.appendleft(beginWord)
# 单词长度
word_n = len(beginWord)
res = 1
# BFS
while queue:
n = len(queue)
for _ in range(n):
tmp = queue.pop()
# 等于endWord
if tmp == endWord:
return res
# 访问过的删除
if tmp in wordList:
wordList.remove(tmp)
for i in range(word_n):
for a in range(97, 123):
w = tmp[:i] + chr(a) + tmp[i+1:]
#print(w)
if w in wordList:
queue.appendleft(w)
res += 1
return 0
上面过了30/40(只多过一个),哈哈,还是不行!再优化, 换一个BFS框架,
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
from collections import deque
wordList = set(wordList)
if endWord not in wordList:
return 0
cur = set()
cur.add(beginWord)
# 单词长度
word_n = len(beginWord)
res = 1
# BFS
while cur:
next_time = set()
if endWord in cur:
return res
for tmp in cur:
if tmp in wordList:
wordList.remove(tmp)
for i in range(word_n):
for a in range(97, 123):
w = tmp[:i] + chr(a) + tmp[i+1:]
if w in wordList:
next_time.add(w)
res += 1
cur = next_time
return 0
终于经过各种操作, 终于可以过了, 但是速度还是太慢
看到一个技巧, 就是beginWord和endWord一起走, 什么意思呢?就是说beginWord找endWord,endWord找beginWord,直接看代码也很好理解的!
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
if endWord not in wordList:
return 0
wordict = set(wordList)
s1 = {beginWord}
s2 = {endWord}
n = len(beginWord)
step = 0
wordict.remove(endWord)
while s1 and s2:
step += 1
if len(s1) > len(s2): s1, s2 = s2, s1
s = set()
for word in s1:
nextword = [word[:i] + chr(j) + word[i + 1:] for j in range(97, 123) for i in range(n)]
for w in nextword:
if w in s2:
return step + 1
if w not in wordict: continue
wordict.remove(w)
s.add(w)
s1 = s
return 0
java
class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> wordSet = new HashSet<>(wordList.size());
wordSet.addAll(wordList);
if (!wordSet.contains(endWord)) return 0;
Set<String> s1 = new HashSet<>();
Set<String> s2 = new HashSet<>();
s1.add(beginWord);
s2.add(endWord);
int n = beginWord.length();
int step = 0;
while (!s1.isEmpty() && !s2.isEmpty()) {
step++;
if (s1.size() > s2.size()) {
Set<String> tmp = s1;
s1 = s2;
s2 = tmp;
}
Set<String> s = new HashSet<>();
for (String word : s1) {
for (int i = 0; i < n; i++) {
char[] chars = word.toCharArray();
for (char ch = 'a'; ch <= 'z'; ch++) {
chars[i] = ch;
String tmp = new String(chars);
if (s2.contains(tmp)) return step + 1;
if (!wordSet.contains(tmp)) continue;
wordSet.remove(tmp);
s.add(tmp);
}
}
}
s1 = s;
}
return 0;
}
}
