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Topic describes
Define a function that takes the head node of a linked list, reverses the list, and outputs the head node of the reversed list.
Example:
Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULLCopy the code
Solution 1: Iteration (double pointer)
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This method is to traverse the linked list, and then change the direction of next when visiting each node to achieve the purpose of reversing the linked list.
- Initialize the cur and pre nodes, pointing to head and NULL, respectively.
- Loop through the linked list and declare the temp node to hold the next node of the current node.
- Change the next pointer to the current cur node to the pre node.
- Change the pre node to a CUR node.
- Change the cur node to temp.
- Continue processing until the cur node is null and the pre node is returned.
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; *} * /
/ * * *@param {ListNode} head
* @return {ListNode}* /
const reverseList = (head) = > {
let cur = head; // The head pointer to the forward list
let pre = null; // The head pointer to the reverse list
while (cur) {
const temp = cur.next; // Hold the subsequent nodes of the current node to update the forward list
cur.next = pre; // Point the current node to the reverse list. This is the process of establishing a back link
pre = cur; // Update the head pointer of the reverse list to the currently processed node. The round of the reverse list is completed
cur = temp; // Replace the forward list header pointer with a temporary node. The forward list processing is complete and the next round of processing begins
}
return pre;
};
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Complexity analysis
- Time complexity O(N) : Traversing lists uses linear size time.
- Spatial complexity O(1) : The variables pre and cur use constant size for extra space.
Solution two: recursion
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When dealing with a list recursively, you start at the first node of the list, find the last node, which is the head of the reverse list, and go back.
- The head identifier of the initial linked list.
- Return head if head is empty or head.next is empty.
- Define the reverseHead node to hold the reversed list values.
- Each time, the next node of head points to head, creating a reversal.
- Recursive to the last node, return to the reverseHead.
/** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; *} * /
/ * * *@param {ListNode} head
* @return {ListNode}* /
const reverseList = (head) = > {
// Check whether the current node needs processing
if (head == null || head.next == null) {
return head;
}
// Recursively process subsequent nodes
const reverseHead = reverseList(head.next);
// local reverse node
head.next.next = head;
head.next = null;
return reverseHead;
};
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Complexity analysis:
- Time complexity O(N) : N is the length of the linked list, which requires the inversion of each node in the list.
- Spatial complexity O(N) : N is the length of the linked list. The spatial complexity depends mainly on the stack space of recursive calls, which is at most N layers.
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