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.
遞歸
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *reverseList( ListNode *head){
if( head == NULL || head->next == NULL)
return head;
ListNode *p = reverseList( head->next);
head->next->next = head;
head->next = NULL;
return p;
}
ListNode *reverseBetween(ListNode *head, int m, int n) {
if( m >= n)
return head;
ListNode dummy(-1);
ListNode *p = head, *q = head, *pre_p = &dummy;
pre_p->next = head;
while( m > 1){
pre_p = p;
p = p->next;
--m;
}
while( n > 1){
q = q->next;
--n;
}
ListNode *tmp = q->next;
q->next = NULL;
ListNode *head2 = reverseList( p);
pre_p->next = q;
p->next = tmp;
return dummy.next;
}
};優雅的迭代
/**
* 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) {
ListNode dummy(-1);
dummy.next = head;
ListNode *prev = &dummy;
for (int i = 0; i < m-1; ++i)
prev = prev->next;
ListNode* const head2 = prev;
prev = head2->next;
ListNode *cur = prev->next;
for (int i = m; i < n; ++i) {
prev->next = cur->next;
cur->next = head2->next;
head2->next = cur; // 頭插法
cur = prev->next;
}
return dummy.next;
}
};