LeetCode | Odd Even Linked List

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

建两个dummy node,一个维护奇数节点构成的链表,另一个维护偶数节点构成的链表,最后再拼接起来。注意偶数节点构成的链表最后一个节点的next域要置空,否则可能会导致出现环路。

class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode dummy_odd(0), dummy_even(0);
        ListNode *p1 = &dummy_odd, *p2 = &dummy_even;
        bool is_odd = true;
        while (head) {
            if (is_odd) { p1->next = head; p1 = p1->next;}
            else { p2->next = head; p2 = p2->next; }
            head = head->next;
            is_odd = !is_odd;
        }
        p1->next = dummy_even.next;
        p2->next = NULL;  // very important
        return dummy_odd.next;
    }
};
原文地址:https://www.cnblogs.com/ilovezyg/p/6379809.html