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【LeetCode】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。
主要就是考虑边界条件。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */struct Node{ ListNode* head; ListNode* tail; Node(ListNode* begin, ListNode* end): head(begin), tail(end) {}};class Solution {public: ListNode *reverseBetween(ListNode *head, int m, int n) { //find begin ListNode* tail = head; ListNode* cur = head; ListNode* begin = NULL; ListNode* end = NULL; ListNode* pre = NULL; int tmp = m; while(--tmp)//--tmp { pre = cur; cur = cur->next; } begin = cur; tmp = n-m; while(tmp--)//tmp-- cur = cur->next; end = cur; while(tail->next != NULL) tail = tail->next; //4 cases if(begin == head) {//result->head is head if(end == tail) { //reverse all, return result->head Node* result = reverse(begin, end); result->tail->next = NULL; return result->head; } else { //reverse from head to end, end->next=post, and return result->head ListNode* post = end->next; end->next = NULL; Node* result = reverse(begin, end); result->tail->next = post; return result->head; } } else {//head is head if(end == tail) { //reverse from begin to tail, pre->next = result->head, and return head Node* result = reverse(begin, end); pre->next = result->head; result->tail->next = NULL; return head; } else { //reverse from begin to end, pre->next = begin, end->next = post, and return head ListNode* post = end->next; end->next = NULL; Node* result = reverse(begin, end); pre->next = result->head; result->tail->next = post; return head; } } } Node* reverse(ListNode* begin, ListNode* end) { Node* result = new Node(begin, end); if(begin == end) return result; //no change else if(begin->next == end) { end->next = begin; begin->next = NULL; result->head = end; result->tail = begin; return result; } else { ListNode* pre = begin; ListNode* cur = pre->next; ListNode* post = cur->next; while(post != NULL) { cur->next = pre; pre = cur; cur = post; post = post->next; } cur->next = pre; begin->next = NULL; result->head = cur; result->tail = begin; return result; } }};
【LeetCode】Reverse Linked List II
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