Description
Given the head of a singly linked list, reverse the list, and return the reversed list.
Example 1:
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| Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
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Example 2:
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| Input: head = [1,2]
Output: [2,1]
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Example 3:
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| Input: head = []
Output: []
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Constraints:
- The number of nodes in the list is the range
[0, 5000]. -5000 <= Node.val <= 5000
Solutions
Assume the linked list is (1→2→3→∅), and we want to change it to (∅←1←2←3).
While traversing the linked list, change the next pointer of the current node to point to the previous node. Since a node does not have a reference to its previous node, we must store the previous node beforehand. We also need to store the next node before changing the reference. Finally, return the new head reference.
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| /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* slow = nullptr;
ListNode* current = head;
while (current) {
ListNode* next = current->next;
current->next = slow;
slow = current;
current = next;
}
return slow;
}
};
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