【算法设计与分析作业题】第十五周:24. Swap Nodes in Pairs
题目
C++ solution
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if (head == NULL || head->next == NULL)
return head;
ListNode* first = head;
ListNode* second = head->next;
first->next = second->next;
second->next = first;
head = second;
ListNode* last = first;
first = first->next;
if (first == NULL)
return head;
second = first->next;
while (second != NULL) {
first->next = second->next;
second->next = first;
last->next = second;
last = first;
first = first->next;
if (first == NULL)
break;
second = first->next;
}
return head;
}
};
简要题解
按链表顺序交换每两个结点的位置,仔细进行指针操作,注意指针越界问题,即可求解。