Given a linked list, swap every two adjacent nodes and return its head.
You may not modify the values in the list’s nodes, only nodes itself may be changed.
大概意思是说每两个节点为一组交换位置。只使用常量空间,不能修改链表的值,只能修改链表的指针
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode *dummy = new ListNode(-1), *pre = dummy;
dummy->next = head;
while (pre->next && pre->next->next) {
ListNode *t = pre->next->next;
pre->next->next = t->next;
t->next = pre->next;
pre->next = t;
pre = t->next;
}
return dummy->next;
}
};
该代码已经是非常经典了,但乍一看还是有点晕,我们来对代码进行一步步的解析。
可以看出该代码实现了两个相邻节点的交换,最后并将pre指向了下一个节点,然后继续该过程
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if (!head || !head->next) return head;
ListNode *t = head->next;
head->next = swapPairs(head->next->next);
t->next = head;
return t;
}
};
Leetcode 24. Swap Nodes in Pairs(详细图解一看就会)
原文:https://www.cnblogs.com/PixelOrange/p/13552972.html