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.
解法一:递归
/** * 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) return head; ListNode* newhead = head->next; head->next = swapPairs(newhead->next); newhead->next = head; return newhead; } };
解法二:Reverse Nodes in k-Group令k=2
/** * 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) { return reverseKGroup(head,2); } ListNode *reverseKGroup(ListNode *head, int k) { ListNode* newhead = new ListNode(-1); ListNode* tail = newhead; ListNode* begin = head; ListNode* end = begin; while(true) { int count = k; while(count && end != NULL) { end = end->next; count --; } if(count == 0) {//reverse from [begin, end) stack<ListNode*> s; while(begin != end) { s.push(begin); begin = begin->next; } while(!s.empty()) { ListNode* top = s.top(); s.pop(); tail->next = top; tail = tail->next; } } else {//leave out tail->next = begin; break; } } return newhead->next; } };
解法三:
每两个节点成对交换次序后,返回给前一个结点进行连接。
/** * 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* newhead = new ListNode(-1); newhead->next = head; ListNode* tail = newhead; while(tail->next != NULL && tail->next->next != NULL) { ListNode* A = tail->next; ListNode* B = A->next; //swap A->next = B->next; B->next = A; tail->next = B; tail = tail->next->next; } return newhead->next; } };