Subject description:
Given a singly linked list of nodes head head, and two integers from and to, on the way from the list of the node to the second node to the inverting portion.
E.g:
1-->2-->3-->4-->5-->6-->null，from=3，to=5
Adjustment result is: 1 -> 2 -> 5 -> 4 -> 3 -> 6 -> null
1-->2-->3-->null，from=1，to=3
Adjustment result is: 3 -> 2 -> 1 -> null
Claim:
1, if the chain length is N, the time complexity requirements of O (N), additional space requirements complexity O (1).
2, if 1 <= from not satisfied <= to <= N, then no adjustment.
Topic analysis and problem-solving ideas:
When partial list of inverting, 1 -> 2 -> 3 -> 4 -> 5 -> null, from = 2, to = 4. We know from-1 fPre node, and a node to + tPos at 1, so that part of the list reversing process is as follows:
At this point the list of fPre 1, pre = fPre.next, the next node is assigned a node pre, pre next node is assigned cur, then the time reversed portion of the list on the list and the whole trans turn the same procedure, the only difference is the pre = tPos.
Problem-solving approach is to first find the next node to the previous node and from the. Then from the node to be reversed at. Finally, the judgment is: if fPre = None, then return to the first node to the new list, or else return to the original head node.
Code:
1 def reversePart(head, frm, to): 2 cur = head 3 fPre = None 4 tPos = None 5 length = 0 6 while cur: 7 length += 1 8 if length == frm - 1: 9 fPre = cur 10 else: 11 fPre = fPre 12 if length == to + 1: 13 tPos = cur 14 else: 15 tPos = tPos 16 cur = cur.next 17 if frm > to or frm < 1 or to > length: 18 return head 19 if fPre: 20 pre = fPre.next 21 else: 22 pre = head 23 cur = pre.next 24 pre.next = tPos 25 while cur != tPos: 26 next_ = cur.next 27 cur.next = pre 28 pre = cur 29 cur = next_ 30 if fPre: 31 fPre.next = pre 32 return head 33 return pre