leetcode92_反转链表 ||

思路

1. 分别记录left前一个节点，right后一个节点
2. 反转[left,right]区间的链表
3. 拼接

func reverseBetween(head *ListNode, left int, right int) *ListNode {
    count := 1
    cur := head
    
    // pre 用于记录left前一个节点
    // post 用于记录 right 后一个节点
    // subHead 用于记录 待反转的链表区间的头节点
    subHead, pre, post := (*ListNode)(nil), (*ListNode)(nil), (*ListNode)(nil)
    
    for count <= right {
        if count == left {
            subHead = cur
        }
        if count < left {
            pre = cur
        }
        if count == right {
            post = cur.Next
            cur.Next = nil
            break
        }
        cur = cur.Next
        count++
    }

    newHead, newTail := reverse(subHead)
    
    if pre != nil {
        pre.Next = newHead
        newTail.Next = post
        return head
    }

    newTail.Next = post
    return newHead
}

func reverse(head *ListNode) (newHead *ListNode, newTail *ListNode){
    if head == nil || head.Next == nil {
        return head, head
    }
    newTail = head

    pre, cur, next := (*ListNode)(nil), head, head
    for cur.Next != nil {
        next = cur.Next
        cur.Next = pre
        pre = cur
        cur = next
    }
    cur.Next = pre
    newHead = cur
    return 
}