LeetCode 24 - Swap Nodes in Pairs
一、问题描述
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 as2->1->4->3
.
Note: Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
给一个链表,交换每两个相邻的结点,返回新链表。
二、解题报告
解法一:操作值域
直接交换结点的val
是最简单的:
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == NULL)
return NULL;
ListNode* first = head;
ListNode* second = head->next;
while(first!=NULL && second!=NULL) {
int temp = first->val; // 交换val
first->val = second->val;
second->val = temp;
if(first->next!=NULL)
first = first->next->next;
if(second->next!=NULL)
second = second->next->next;
}
return head;
}
};
解法二:操作结点
题目要求:You may not modify the values in the list, only nodes itself can be changed. 好吧,那就来操作结点吧。
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if(head == NULL)
return NULL;
ListNode* first = head;
ListNode* second = head->next;
ListNode* p = new ListNode(0);
head = p;
while(first!=NULL && second!=NULL) {
ListNode* temp1 = first;
ListNode* temp2 = second;
if(second->next!=NULL)
second = second->next->next;
if(first->next!=NULL)
first = first->next->next;
p->next = temp2;
p->next->next = temp1;
p = p->next->next;
p->next = NULL;
}
if(first!=NULL) {
p->next = first;
}
return head->next;
}
};
空间复杂度为
LeetCode答案源代码:https://github.com/SongLee24/LeetCode