Java for LeetCode 127 Word Ladder
2024-10-01 04:16:55
Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, such that:
- Only one letter can be changed at a time
- 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
.
解题思路一:
对set进行逐个遍历,递归实现,JAVA实现如下:
static public int ladderLength(String beginWord, String endWord,
Set<String> wordDict) {
int result = wordDict.size() + 2;
Set<String> set = new HashSet<String>(wordDict);
if (oneStep(beginWord, endWord))
return 2;
for (String s : wordDict) {
if (oneStep(beginWord, s)) {
set.remove(s);
int temp = ladderLength(s, endWord, set);
if (temp != 0)
result = Math.min(result, temp + 1);
set.add(s);
}
}
if (result == wordDict.size() + 2)
return 0;
return result;
} public static boolean oneStep(String s1, String s2) {
int res = 0;
for (int i = 0; i < s1.length(); i++)
if (s1.charAt(i) != s2.charAt(i))
res++;
return res == 1;
}
结果TLE
解题思路二:
发现直接遍历是行不通的,实际上如果使用了oneStep函数,不管怎么弄都会TLE的(貌似在C++中可以AC)。
本题的做法应该是采用图的BFS来做,同时oneStep的匹配也比较有意思,JAVA实现如下:
static public int ladderLength(String start, String end, Set<String> dict) {
HashMap<String, Integer> disMap = new HashMap<String, Integer>();
LinkedList<String> queue = new LinkedList<String>();
queue.add(start);
disMap.put(start, 1);
while (!queue.isEmpty()) {
String word = queue.poll();
for (int i = 0; i < word.length(); i++) {
for (char ch = 'a'; ch <= 'z'; ch++) {
StringBuilder sb = new StringBuilder(word);
sb.setCharAt(i, ch);
String nextWord = sb.toString();
if (end.equals(nextWord))
return disMap.get(word) + 1;
if (dict.contains(nextWord) && !disMap.containsKey(nextWord)) {
disMap.put(nextWord, disMap.get(word) + 1);
queue.add(nextWord);
}
}
}
}
return 0;
}
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