0092-leetcode算法实现之反转链表II-reverseLinkedListII-python&golang实现

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

示例 1：

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
示例 2：

输入：head = [5], left = 1, right = 1
输出：[5]

提示：

链表中节点数目为 n
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n

进阶： 你可以使用一趟扫描完成反转吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-linked-list-ii

python

# 0092.反转链表II
# https://leetcode-cn.com/problems/reverse-linked-list-ii/solution/java-shuang-zhi-zhen-tou-cha-fa-by-mu-yi-cheng-zho/
class ListNode:
    def __init__(self, val):
        self.val = val
        self.next = None

class Solution:
    def reverseLinkedListII(self, head: ListNode, left: int, right: int) -> ListNode:
        """
        双指针， 头插法
        :param head:
        :param left:
        :param right:
        :return:
        """
        dummyHead = ListNode(0)
        dummyHead.next = head
        # 初始化
        guard = dummyHead
        p = dummyHead.next
        # 指针移动相应位置
        for i in range(0, left-1):
            guard = guard.next
            p = p.next
        #
        for i in range(0, right-left):
            removed = p.next
            p.next = p.next.next
            removed.next = guard.next
            guard.next = removed

        retNode = dummyHead.next
        dummyHead.next = None
        return retNode

golang

// 双指针
func reverseLinkedListII(head *ListNode, left, right int) *ListNode {
	dummyHead := &ListNode{}
	dummyHead.Next = head

	guard := dummyHead
	p := dummyHead.Next

	for i := 0; i < left-1; i++ {
		guard = guard.Next
		p = p.Next
	}

	for i := 0; i < right-left; i++ {
		removed := p.Next
		p.Next = p.Next.Next
		removed.Next = guard.Next
		guard.Next = removed
	}
	return dummyHead.Next
}