leetcode - Reverse Linked List II
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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.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */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 *p = head;std::vector<int> vec;while(p != NULL){vec.push_back(p->val);p = p->next;}while(m < n){std::swap(vec[m-1],vec[n-1]);m++;n--;}ListNode *node = new ListNode(0);ListNode *p1 = node;ListNode *p2 = NULL;for (int i = 0; i < vec.size(); i++){p2 = new ListNode(vec[i]);p1->next = p2;p1 = p2;}return node->next; }};
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