86. 分隔链表

原题

https://leetcode.cn/problems/partition-list/description/

思路

双指针

复杂度

时间：O(n)
空间：O(1)

Python代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
        # 虚拟节点, p1指向小节点, p2指向大节点
        dummy1 = p1 = ListNode(-1, head)
        dummy2 = p2 = ListNode(-1, head)
        while head:
            if head.val < x:
                p1.next = head 
                p1 = p1.next 
            else:
                p2.next = head 
                p2 = p2.next 
            head = head.next 
        p2.next = None
        p1.next = dummy2.next 
        return dummy1.next

Go代码

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func partition(head *ListNode, x int) *ListNode {
	// 虚拟节点, p1指向小节点, p2指向大节点
	dummy1 := &ListNode{-1, head}
	dummy2 := &ListNode{-1, head}
	p1 := dummy1
	p2 := dummy2
	for head != nil {
		if head.Val < x {
			p1.Next = head
			p1 = p1.Next
		} else {
			p2.Next = head
			p2 = p2.Next
		}
		head = head.Next
	}
	p2.Next = nil
	p1.Next = dummy2.Next
	return dummy1.Next
}