算法:
对于链表的排序,一般要设计到拆分合并两步,拆分这一步:
中间节点作为临界值,小的放左边,大的放右边
合并操作步骤:
将两个有序的链表中,串联起来
题目1:分隔链表
https://leetcode-cn.com/problems/partition-list/submissions/
代码实现:
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func partition(head *ListNode, x int) *ListNode {
if head == nil {
return nil
}
curr := head
before := new(ListNode)
before1 := before
after := new(ListNode)
after1 := after
for curr != nil {
if curr.Val < x {
before.Next = curr
before = before.Next
} else {
after.Next = curr
after= after.Next
}
curr = curr.Next
}
before.Next = nil
after.Next = nil
if after1.Next != nil {
before.Next = after1.Next // after1记录偏移之前的after首节点位置
}
return before1.Next // 原因是before1首节点是一个none的节点。
}
/* 解法:
这个可以拆分成,两个链表,小于x的放到before,大于等于的放到after.
然后将这两个链表拼接起来。
*/
执行结果:
题目2: https://leetcode-cn.com/problems/partition-list-lcci/submissions/
代码实现:
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func partition(head *ListNode, x int) *ListNode {
l1, l2 := new(ListNode),new(ListNode)
res,res1 := l1,l2
for head != nil {
if head.Val < x {
l1.Next = &ListNode{Val:head.Val}
l1 = l1.Next
} else {
l2.Next = &ListNode{Val:head.Val}
l2 = l2.Next
}
head = head.Next
}
l1.Next = res1.Next
return res.Next
}
// 双指针排序,小于x的放到l1,大于x的放在l2; 最后将两个链表串起来
执行结果:
题目3:排序链表
https://leetcode-cn.com/problems/sort-list/
代码实现:
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func sortList(head *ListNode) *ListNode {
// 归并写法:1.先拆分,二分法拆分,快慢指针;2.在合并,双指针的方式
if head == nil || head.Next == nil {
return head
}
// 快慢指针找到对应的中间位置节点
s,f := head,head.Next // 两个指针不要指向同一个节点
for f != nil && f.Next != nil {
s = s.Next
f = f.Next.Next
}
tmp := s.Next
s.Next = nil
// 递归操作左右链表
l := sortList(head)
r := sortList(tmp)
pre := new(ListNode)
res := pre
// 合并左右链表
for l != nil && r != nil {
if l.Val < r.Val {
pre.Next = l
l = l.Next
} else {
pre.Next = r
r = r.Next
}
pre = pre.Next
}
if l != nil {
pre.Next = l
} else if r != nil {
pre.Next = r
}
return res.Next
}
执行结果: