【LeetCode】92. Reverse Linked List II

Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given

1->2->3->4->5->NULL

, m = 2 and n = 4,

return

1->4->3->2->5->NULL

.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

把[m,n]那一段抠出来，reverse之后，再拼回去。

需要注意的是，reverse函数的参数需要是指针引用*&。

附：指针传递与指针引用传递的区别

当我们把指针做为参数传递时，其实是把指针的副本传递给了函数，也可以说传递指针是指针的值传递。这样当我们在函数内部修改指针时，在函数里修改只是修改的指针的副本，而不是指针本身，原来的指针还保留着原来的值。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
if(head == NULL)
return head;

ListNode *newhead = new ListNode(-1);
newhead->next = head;

ListNode *prebegin = newhead;
ListNode *begin = head;

ListNode *end = newhead;
ListNode *postend = head;

while(--m)
{
prebegin = prebegin->next;
begin = begin->next;
}
while(n--)
{
end = end->next;
postend = postend->next;
}
//reverse
reverse(begin, end);
//link
prebegin->next = begin;
end->next = postend;

return newhead->next;
}
void reverse(ListNode*& begin, ListNode*& end)
{
if(begin == end)
return;
else if(begin->next == end)
end->next = begin;
else
{//at least 3 nodes
ListNode* pre = begin;
ListNode* cur = pre->next;
ListNode* post = cur->next;
while(post != end->next)
{
cur->next = pre;
pre = cur;
cur = post;
post = post->next;
}
cur->next = pre;
}
//swap
ListNode* temp = begin;
begin = end;
end = temp;
}
};