【LeetCode】328. Odd Even Linked List 解題報告

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328. Odd Even Linked List / Medium

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]

Constraints:

  • The number of nodes in the linked list is in the range [0, 10^4].
  • -10^6 <= Node.val <= 10^6

Solution:

思路

題目直接告訴我們要在 in-place 執行，而且時間複雜度要 O(n)，
但解法也很簡單，就是直接將基數跟偶數節點拆開。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(1)

LeetCode Result

Code

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class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode *head_even = head->next; 
        ListNode *head_odd = head;
        ListNode *cur_even = head_even;
        ListNode *cur_odd = head_odd;
        
        while(cur_odd->next && cur_even->next) {
            cur_odd->next = cur_even->next;
            cur_odd = cur_odd->next;
            cur_even->next = cur_odd->next;
            cur_even = cur_even->next;
        }
        
        cur_odd->next = head_even;
        return head_odd;
    }
};