328. Odd Even Linked List

Leetcode

https://leetcode.com/problems/odd-even-linked-list/

題目

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

Example 1:

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2:

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

解答

• 方法一

var oddEvenList = function(head) {
    if(!head) return null;
    if(!head.next || !head.next.next) return head;
    
    let pointer = head;
    let lastNode = head.next;
    const evenStartNode = lastNode;
    while(lastNode && lastNode.next) {
        const oddNode = lastNode.next;

        lastNode.next = oddNode.next;
        lastNode = oddNode.next;
        
        pointer.next = oddNode;
        oddNode.next = evenStartNode;
        pointer = pointer.next;
    }
    return head;
};

Runtime: 84 ms, faster than 76.58% of JavaScript online submissions for Odd Even Linked List.

Memory Usage: 41.1 MB, less than 59.92% of JavaScript online submissions for Odd Even Linked List.

For example:

head = [1, 2, 3, 4, 5, 6, 7] -> pointer = 1， lastNode = 2，evenStartNode = 2

1. oddNode = 3， 2 -> 4，lastNode = 4，1 -> 3，3 -> 2，pointer = 3

result: 1 -> 3 -> 2 -> 4 -> 5 -> 6 -> 7，pointer = 3，lastNode = 4

2. oddNode = 5，4 -> 6，lastNode = 6，3 -> 5 -> 2，pointer = 5

result: 1 -> 3 -> 5 -> 2 -> 4 -> 6 -> 7，pointer = 5，lastNode = 6

3. oddNode = 7，6 -> null，lastNode = null，5 -> 7 -> 2，pointer = 7

result: 1 -> 3 -> 5 -> 7 -> 2 -> 4 -> 6 -> null，pointer = 7，lastNode = null