Swap Nodes in Pairs
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.
//测试Accepted //要注意的一点是一旦是把A和B交换了以后,A的父节点就指向了B。这个是容易忘记的一点。 class Solution { public: ListNode *swapPairs(ListNode *head) { ListNode* prev = NULL; ListNode* first = NULL; ListNode* second = head; int k = 1; while (second != NULL) { if (k%2 == 0) { first->next = second->next; second->next = first; if(prev != NULL)//when process prev pointer, we should be very careful prev->next = second; else head = second; ListNode* tmp = first; first = second; second = tmp; } prev = first; first = second; second = second->next; k++; } return head; } };
//要注意的一点是一旦是把A和B交换了以后,A的父节点就指向了B。这个是容易忘记的一点。 #include<iostream> using namespace std; #define N 5 struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *swapPairs(ListNode *head) { ListNode* prev = NULL; ListNode* first = NULL; ListNode* second = head; int k = 1; while (second != NULL) { if (k%2 == 0) { first->next = second->next; second->next = first; if(prev != NULL)//when process prev pointer, we should be very careful prev->next = second; else head = second; ListNode* tmp = first; first = second; second = tmp; } prev = first; first = second; second = second->next; k++; } return head; } }; ListNode *creatlist() { ListNode *head; head=NULL; for(int i=0; i<N; i++) { int a; ListNode *p; cin>>a; p = (ListNode*)malloc(sizeof(ListNode)); p->val=a; p->next=head; head = p; } return head; } int main() { ListNode *list=creatlist(); Solution lin; ListNode *outlist=lin.swapPairs(list); for(int i=0; i<N; i++) { cout<<outlist->val; outlist = outlist->next; } }
leetcode_24_Swap Nodes in Pairs
原文地址:http://blog.csdn.net/keyyuanxin/article/details/43791899