【題目】
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.
【分析】
使用遞歸解決,首先跳出遞歸的條件一定是鏈表剩餘長度爲0或1,即head == NULL,head->next == NULL. 其次,一定要注意在鏈表還有剩餘節點的情況下倒轉兩個節點,不然,head->next都爲空了,如何反轉後面節點呢?具體見下面註釋。
【代碼】
運行時間4ms
/**
* 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) // 遞歸跳出條件
return head;
ListNode* tmp = swapPairs(head->next->next); //一定要在前面遞歸,如果head->next都爲空了,遞歸進去是可以看到的,但是下面直接去head->next就判斷不了。
ListNode* res = head->next;
res->next = head;
res->next->next = tmp;
return res;
}
};