标签:dict rdl int output The word hang begin before
Given two words beginWord
and endWord
, and a dictionary wordList
, return the length of the shortest transformation sequence from beginWord
to endWord
, such that:
Return 0
if there is no such transformation sequence.
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.
Constraints:
1 <= beginWord.length <= 100
endWord.length == beginWord.length
1 <= wordList.length <= 5000
wordList[i].length == beginWord.length
beginWord
, endWord
, and wordList[i]
consist of lowercase English letters.beginWord != endWord
wordList
are unique.给定一个初始字符串、目标字符串、一组字符串序列,从初始字符串开始,每次改变其中一个字符,使得到的新字符串是序列中的一个,重复操作,判断最终能否得到目标字符串。
BFS。从初始字符串开始,变动其中的每一个字符,将其改为另外25个字母,判断得到的新字符串是否在序列中,如果在的话加入队中,重复操作直到找到目标字符串。
class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordList) {
Set<String> set = new HashSet<>();
Queue<String> q = new LinkedList<>();
int step = 0;
for (String word : wordList) {
set.add(word);
}
q.offer(beginWord);
while (!q.isEmpty()) {
step++;
int size = q.size();
for (int i = 0; i < size; i++) {
String cur = q.poll();
for (int index = 0; index < cur.length(); index++) {
char c = cur.charAt(index);
String before = cur.substring(0, index);
String after = cur.substring(index + 1, cur.length());
for (int j = 0; j < 26; j++) {
if (j != c - ‘a‘) {
String newWord = before + (char) (‘a‘ + j) + after;
if (set.contains(newWord)) {
if (endWord.equals(newWord)) {
return step + 1;
}
q.offer(newWord);
set.remove(newWord);
}
}
}
}
}
}
return 0;
}
}
标签:dict rdl int output The word hang begin before
原文地址:https://www.cnblogs.com/mapoos/p/14256677.html