目录 
题目
Reverse a linked list from position m to n. Do it in-place and in one-pass.
Example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Example:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
解决方案
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* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if(head == null) return null;
int tempA = -1;
ListNode pre = null;
ListNode begin = null;
ListNode end = null;
ListNode before = null;
ListNode fakeHead = new ListNode(0);
fakeHead.next = head;
ListNode p = fakeHead;
while(p != null){
tempA++;
if(tempA < m-1){
p = p.next;
}
if((tempA+1) == m) {
pre = p;
before = p;
p = p.next;
}
if(tempA == m) {
begin = p;
}
if(m <= tempA && tempA <= n){
ListNode next = p.next;
p.next = pre;
pre = p;
if(tempA == n){
end = p;
}
p = next;
}
if(tempA == n){
begin.next = p;
if(before != null){
before.next = end;
}
break;
}
}
return fakeHead.next;
}
}
|
注意事项
- 使用“头指针”，从而去除头结点的特殊性。
- 难度在于要只一遍遍历且原地完成。
- 记录要翻转节点的前一个节点、要翻转的第一个节点、要翻转的最后一个节点。
- 核心代码：
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| if(m <= tempA && tempA <= n){
ListNode next = p.next;
p.next = pre;
pre = p;
if(tempA == n){
end = p;
}
p = next;
}
|
是使用了链表翻转的典型做法。
