[Leetcode] #92#206 Reverse Linked List I & II

Discription:

Reverse a singly linked list.

Solution:

ListNode* reverseList(ListNode* head) {
ListNode *pre = NULL, *cur = head;
while (cur){
ListNode *temp = cur->next;
cur->next = pre;
pre = cur;
cur = temp;
}
return pre;
}

ListNode* reverseList(ListNode* head) {
if (head == NULL) return NULL;
ListNode *pNext, *pPre = NULL;
while (head->next){
pNext = head->next;
head->next = pPre;
pPre = head;
head = pNext;
}
head->next = pPre;
return head;
}

ListNode* reverseList(ListNode* head) {
if (!head || !(head->next)) return head;
ListNode* node = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return node;
}

Discription:

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

.

Solution:

ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode *pre = new ListNode(-1);
pre->next = head;
head = pre;
ListNode *p1 = head, *p2 = head;
while (m != 0 && p1->next != NULL){
pre = p1;
p1 = p1->next;
m--;
}
while (n != 0 && p2->next != NULL){
p2 = p2->next;
n--;
}
ListNode *pPre = p2->next;
while (p1 != p2){
ListNode *pNext = p1->next;
p1->next = pPre;
pPre = p1;
p1 = pNext;
}
p2->next = pPre;
pre->next = p2;
return head->next;
}

ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode* new_head = new ListNode(0);
new_head->next = head;
ListNode* pre = new_head;
for (int i = 0; i < m - 1; i++)
pre = pre->next;
ListNode* cur = pre->next;
for (int i = 0; i < n - m; i++) {
ListNode* move = cur->next;
cur->next = move->next;
move->next = pre->next;
pre->next = move;
}
return new_head->next;
}

GitHub-LeetCode: https://github.com/wenwu313/LeetCode