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 + } +} +``` -----------------------
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 878b2466..8fd9c604 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 @@ -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+=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 + } +} +``` + + C#: ```csharp public bool IsAnagram(string s, string t) { @@ -326,6 +352,7 @@ C#: return true; } ``` + ## 相关题目 * 383.赎金信