Merge pull request #1315 from wzqwtt/patch05

添加(0142.环形链表II、0242.有效的字母异位词)Scala版本
This commit is contained in:
程序员Carl
2022-06-01 08:21:06 +08:00
committed by GitHub
2 changed files with 52 additions and 1 deletions

View File

@ -370,7 +370,31 @@ ListNode *detectCycle(ListNode *head) {
}
```
Scala:
```scala
object Solution {
def detectCycle(head: ListNode): ListNode = {
var fast = head // 快指针
var slow = head // 慢指针
while (fast != null && fast.next != null) {
fast = fast.next.next // 快指针一次走两步
slow = slow.next // 慢指针一次走一步
// 如果相遇fast快指针回到头
if (fast == slow) {
fast = head
// 两个指针一步一步的走,第一次相遇的节点必是入环节点
while (fast != slow) {
fast = fast.next
slow = slow.next
}
return fast
}
}
// 如果fast指向空值必然无环返回null
null
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -308,6 +308,32 @@ impl Solution {
}
```
Scala:
```scala
object Solution {
def isAnagram(s: String, t: String): Boolean = {
// 如果两个字符串的长度不等直接返回false
if (s.length != t.length) return false
val record = new Array[Int](26) // 记录每个单词出现了多少次
// 遍历字符串对于s字符串单词对应的记录+=1t字符串对应的记录-=1
for (i <- 0 until s.length) {
record(s(i) - 97) += 1
record(t(i) - 97) -= 1
}
// 如果不等于则直接返回false
for (i <- 0 until 26) {
if (record(i) != 0) {
return false
}
}
// 如果前面不返回false说明匹配成功返回truereturn可以省略
true
}
}
```
C#
```csharp
public bool IsAnagram(string s, string t) {
@ -326,6 +352,7 @@ C#
return true;
}
```
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