题解 | 链表的奇偶重排

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        // write code here
        int a[100010], b[100010], cnt1 = 0 , cnt2 = 0;
        while(head) 
        {
            a[++ cnt1] = head -> val;
            head = head -> next;
        }

        for(int i = 1; i <= cnt1; i ++) 
            if(i % 2 == 1) b[++ cnt2] = a[i];
        for(int i = 1; i <= cnt1; i ++) 
            if(i % 2 != 1) b[++ cnt2] = a[i];

        ListNode* phead = new ListNode(-1);
        ListNode* cur = phead;
        for(int i = 1; i <= cnt2; i ++) 
        {
            ListNode* p = new ListNode(b[i]);
            cur -> next = p;
            cur = cur -> next;
        }

        return phead -> next;
    }
};