(算法分析Week11)Swap Nodes in Pairs[Medium]

24.Swap Nodes in Pairs[Medium]

题目来源

Description

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.

Solution

对链表进行操作，画图很直观清晰。

这道题第一反应就是遍历整个链表然后对节点进行操作，每次跳两个节点，后一个接到前面，前一个接到后一个的后面，最后现在的后一个（也就是原来的前一个）接到下下个结点（如果没有则接到下一个）。注意边界情况即可。习惯加一个表头，方便操作。

看Discussion时发现递归更简单

if (head == null || head.next == null) {
return head;
}

ListNode newhd = head.next;
head.next = swapPairs(newhd.next);
newhd.next = head;
return newhd;

Complexity analysis

O（n）n为节点数。

Code

/**
* 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) {
ListNode* helper = new ListNode(0);
helper -> next = head;
ListNode* pre = helper;
ListNode* cur = head;
while(cur && cur -> next) {
ListNode* Next = cur -> next -> next;
cur -> next -> next = cur;
pre -> next = cur -> next;
if (Next != NULL && Next -> next != NULL) {
cur -> next = Next -> next;
}
else {
cur -> next = Next;
}
pre = cur;
cur = Next;
}
return helper -> next;
}
};

Result