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/************************************************************************

* * Given a linked list, swap every two adjacent nodes and return its head. * * For example, * Given 1->2->3->4, you should return the list as 2->1->4->3. * * Your algorithm should use only constant space. You may not modify the values in the list, * only nodes itself can be changed. * * ************************************************************************/ 关于交换对节点的解析如下：

完整AC代码

ListNode* swapPairs(ListNode* head) {        ListNode preHead(0), *pre = &preHead;        ListNode *cur = head;        pre->next = head;        while (cur&&cur->next)        {            ListNode* nxt = cur->next;            cur->next = nxt->next;            nxt->next = pre->next;            pre->next = nxt;            pre = cur;            cur = cur->next;        }        return preHead.next;    }

/************************************************************************

* * Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. * * If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. * * You may not alter the values in the nodes, only nodes itself may be changed. * * Only constant memory is allowed. * * For example, * Given this linked list: 1->2->3->4->5 * * For k = 2, you should return: 2->1->4->3->5 * * For k = 3, you should return: 3->2->1->4->5 * * ************************************************************************/

与上一题基本相同，区别就是要知道链表的长度，当剩余链表的长度小于k是，就不需要做了。

AC代码

ListNode *reverseKGroup(ListNode *head, int k) {        if (!head || k == 1) return head;        int listLen = 0;        ListNode preHead(0);        preHead.next = head;        ListNode* cur = &preHead, *pre = &preHead, *next = NULL;        while (cur = cur->next) listLen++;        while (listLen >= k) {            cur = pre->next;            next = cur->next;            for (int i = 0; i < k - 1; i++) {                cur->next = next->next;                next->next = pre->next;                pre->next = next;                next = cur->next;            }            pre = cur;            listLen -= k;        }        return preHead.next;    }