[LeetCode] 088: Reverse Linked List II

[Problem]

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.

[Solution]

class Solution {
public:
ListNode *reverseBetween(ListNode *head, int m, int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
// need not reverse
if(NULL == head || m == n)return head;

// get the position of m and n
ListNode *p1 = head;// node at position m
ListNode *p2 = head;// node at position n
ListNode *pre = head;// node before p1
ListNode *after = head;// node after p2
for(int i = 0; i < n-m; ++i){
p2 = p2->next;
}
for(int i = 1; i < m; ++i){
pre = p1;
p1 = p1->next;
p2 = p2->next;
}
after = p2->next;

// reverse [m:n]
ListNode *pi = p1;
ListNode *pj = pi->next;
ListNode *pk = pj->next;
while(pi != p2){
pj->next = pi;

// update pi, pj, pk
pi = pj;
pj = pk;
if(pk == NULL)break;
pk = pk->next;
}

// link pre->p2 and p1->after
if(pre == p1){
head = p2;
}
else{
pre->next = p2;
}
p1->next = after;

return head;
}
};

说明：版权所有，转载请注明出处。Coder007的博客