algorithm/LeetCode-Go
1
0
Fork 0
mirror of https://github.com/halfrost/LeetCode-Go.git synced 2025-07-05 16:36:41 +08:00
Code Issues Packages Projects Releases Wiki Activity

Fix problem 82 gofmt

Browse Source
This commit is contained in:
YDZ
2020-10-01 09:27:18 +08:00
parent deb2aa1dc1
commit 2af7be3791
2 changed files with 103 additions and 102 deletions
Download Patch File Download Diff File Expand all files Collapse all files

102
leetcode/0082.Remove-Duplicates-from-Sorted-List-II/82. Remove Duplicates from Sorted List II.go
View File

@ -94,63 +94,63 @@ func deleteDuplicates(head *ListNode) *ListNode {
// 双循环简单解法 O(n*m)
func deleteDuplicates3(head *ListNode) *ListNode {
if head == nil {
return head
}
if head == nil {
return head
}
nilNode := &ListNode{Val: 0, Next: head}
head = nilNode
nilNode := &ListNode{Val: 0, Next: head}
head = nilNode
lastVal := 0
for head.Next != nil && head.Next.Next != nil {
if head.Next.Val == head.Next.Next.Val {
lastVal = head.Next.Val
for head.Next != nil && lastVal == head.Next.Val {
head.Next = head.Next.Next
}
} else {
head = head.Next
}
}
return nilNode.Next
lastVal := 0
for head.Next != nil && head.Next.Next != nil {
if head.Next.Val == head.Next.Next.Val {
lastVal = head.Next.Val
for head.Next != nil && lastVal == head.Next.Val {
head.Next = head.Next.Next
}
} else {
head = head.Next
}
}
return nilNode.Next
}
// 双指针+删除标志位，单循环解法 O(n)
func deleteDuplicates4(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
if head == nil || head.Next == nil {
return head
}
nilNode := &ListNode{Val: 0, Next: head}
// 上次遍历有删除操作的标志位
lastIsDel := false
// 虚拟空结点
head = nilNode
// 前后指针用于判断
pre, back := head.Next, head.Next.Next
// 每次只删除前面的一个重复的元素，留一个用于下次遍历判重
// pre, back 指针的更新位置和值比较重要和巧妙
for head.Next != nil && head.Next.Next != nil {
if pre.Val != back.Val && lastIsDel {
head.Next = head.Next.Next
pre, back = head.Next, head.Next.Next
lastIsDel = false
continue
}
nilNode := &ListNode{Val: 0, Next: head}
// 上次遍历有删除操作的标志位
lastIsDel := false
// 虚拟空结点
head = nilNode
// 前后指针用于判断
pre, back := head.Next, head.Next.Next
// 每次只删除前面的一个重复的元素，留一个用于下次遍历判重
// pre, back 指针的更新位置和值比较重要和巧妙
for head.Next != nil && head.Next.Next != nil {
if pre.Val != back.Val && lastIsDel {
head.Next = head.Next.Next
pre, back = head.Next, head.Next.Next
lastIsDel = false
continue
}
if pre.Val == back.Val {
head.Next = head.Next.Next
pre, back = head.Next, head.Next.Next
lastIsDel = true
} else {
head = head.Next
pre, back = head.Next, head.Next.Next
lastIsDel = false
}
}
// 处理 [1,1] 这种删除还剩一个的情况
if lastIsDel && head.Next != nil {
head.Next = nil
}
return nilNode.Next
if pre.Val == back.Val {
head.Next = head.Next.Next
pre, back = head.Next, head.Next.Next
lastIsDel = true
} else {
head = head.Next
pre, back = head.Next, head.Next.Next
lastIsDel = false
}
}
// 处理 [1,1] 这种删除还剩一个的情况
if lastIsDel && head.Next != nil {
head.Next = nil
}
return nilNode.Next
}