why did I get the error of time limit exceeded for the problem in leetcode?
0
$begingroup$
def ladderLength(beginWord: String, endWord: String, wordList: List[String]): Int = {
if(beginWord == null || beginWord.isEmpty || endWord == null ||
endWord.isEmpty || !wordList.contains(endWord)) {
return 0
}
var len = 0
val shadow = new HashSet[String]()
val q = new Queue[String]()
q.enqueue(beginWord)
while(!q.isEmpty) {
len += 1
val q1 = q.clone()
q.clear()
while(!q1.isEmpty) {
val cur = q1.dequeue()
for(i <- 0 until cur.length)
for(c <- 'a' to 'z') {
val mutate = cur.updated(i, c)
if(wordList.contains(mutate) && !shadow.contains(mutate)) {
if(mutate == endWord)
return len+1
q.enqueue(mutate)
shadow.add(mutate)
}
}
}
}
0
}
This is my answer to Problem No. 127 in leetcode. But i got "time limit exceeded". anybody can give any hint on how to optimize the code?
algorithm graph
asked 6 mins ago
kenneth.sunkenneth.sun
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
$begingroup$
def ladderLength(beginWord: String, endWord: String, wordList: List[String]): Int = {
if(beginWord == null || beginWord.isEmpty || endWord == null ||
endWord.isEmpty || !wordList.contains(endWord)) {
return 0
}
var len = 0
val shadow = new HashSet[String]()
val q = new Queue[String]()
q.enqueue(beginWord)
while(!q.isEmpty) {
len += 1
val q1 = q.clone()
q.clear()
while(!q1.isEmpty) {
val cur = q1.dequeue()
for(i <- 0 until cur.length)
for(c <- 'a' to 'z') {
val mutate = cur.updated(i, c)
if(wordList.contains(mutate) && !shadow.contains(mutate)) {
if(mutate == endWord)
return len+1
q.enqueue(mutate)
shadow.add(mutate)
}
}
}
}
0
}
This is my answer to Problem No. 127 in leetcode. But i got "time limit exceeded". anybody can give any hint on how to optimize the code?
algorithm graph
asked 6 mins ago
kenneth.sunkenneth.sun
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
$begingroup$
def ladderLength(beginWord: String, endWord: String, wordList: List[String]): Int = {
if(beginWord == null || beginWord.isEmpty || endWord == null ||
endWord.isEmpty || !wordList.contains(endWord)) {
return 0
}
var len = 0
val shadow = new HashSet[String]()
val q = new Queue[String]()
q.enqueue(beginWord)
while(!q.isEmpty) {
len += 1
val q1 = q.clone()
q.clear()
while(!q1.isEmpty) {
val cur = q1.dequeue()
for(i <- 0 until cur.length)
for(c <- 'a' to 'z') {
val mutate = cur.updated(i, c)
if(wordList.contains(mutate) && !shadow.contains(mutate)) {
if(mutate == endWord)
return len+1
q.enqueue(mutate)
shadow.add(mutate)
}
}
}
}
0
}
This is my answer to Problem No. 127 in leetcode. But i got "time limit exceeded". anybody can give any hint on how to optimize the code?
algorithm graph
asked 6 mins ago
kenneth.sunkenneth.sun
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
def ladderLength(beginWord: String, endWord: String, wordList: List[String]): Int = {
if(beginWord == null || beginWord.isEmpty || endWord == null ||
endWord.isEmpty || !wordList.contains(endWord)) {
return 0
}
var len = 0
val shadow = new HashSet[String]()
val q = new Queue[String]()
q.enqueue(beginWord)
while(!q.isEmpty) {
len += 1
val q1 = q.clone()
q.clear()
while(!q1.isEmpty) {
val cur = q1.dequeue()
for(i <- 0 until cur.length)
for(c <- 'a' to 'z') {
val mutate = cur.updated(i, c)
if(wordList.contains(mutate) && !shadow.contains(mutate)) {
if(mutate == endWord)
return len+1
q.enqueue(mutate)
shadow.add(mutate)
}
}
}
}
0
}
This is my answer to Problem No. 127 in leetcode. But i got "time limit exceeded". anybody can give any hint on how to optimize the code?
algorithm graph
algorithm graph
asked 6 mins ago
kenneth.sunkenneth.sun
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
kenneth.sunkenneth.sun
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
kenneth.sunkenneth.sun
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 mins ago
kenneth.sunkenneth.sun
1
asked 6 mins ago
kenneth.sunkenneth.sun
1
1
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
kenneth.sun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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