From 0a58ef986de4c20c5d69a4b490b478f44b7f3f5c Mon Sep 17 00:00:00 2001
From: ZongqinWang <1722249371@qq.com>
Date: Thu, 12 May 2022 17:35:30 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200142.=E7=8E=AF?=
=?UTF-8?q?=E5=BD=A2=E9=93=BE=E8=A1=A8II.md=20Scala=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0142.环形链表II.md | 26 +++++++++++++++++++++++++-
1 file changed, 25 insertions(+), 1 deletion(-)
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md
index e8ca950d..f8e62d45 100644
--- a/problems/0142.环形链表II.md
+++ b/problems/0142.环形链表II.md
@@ -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
+ }
+}
+```
-----------------------
From a25409d7ec351aaf2e6e513e3e091fc97cd17d60 Mon Sep 17 00:00:00 2001
From: ZongqinWang <1722249371@qq.com>
Date: Thu, 12 May 2022 19:16:20 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200242.=E6=9C=89?=
=?UTF-8?q?=E6=95=88=E7=9A=84=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D?=
=?UTF-8?q?.md=20Scala=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0242.有效的字母异位词.md | 25 +++++++++++++++++++++++
1 file changed, 25 insertions(+)
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index 080166fd..4f0d143e 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -307,6 +307,31 @@ 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+=1,t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true,return可以省略
+ true
+ }
+}
+```
+
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