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.
/** * 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 dummy(INT_MIN); ListNode *tail = &dummy; ListNode *first = head; while (first != NULL) { ListNode *second = first->next; if (second == NULL) break; ListNode* tmp = second->next; second->next = first; tail->next = second; tail = first; first = tmp; } // odd numbers tail->next = first ? first : NULL; return dummy.next; } }; // runtime contribution 25.86%参考后 I. 指针
/** * 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) { typedef ListNode* Node; Node *p = &head; Node first = NULL; Node second = NULL; while ((first = *p) && (second = first->next)) { first->next = second->next; second->next = first; *p = second; p = &(first->next); } return head; } }; // runtime contribution 25.86%II. 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) return NULL; if (head->next == NULL) return head; ListNode* first = head, *second = head->next; first->next = second->next; second->next = first; first->next = swapPairs(first->next); return second; } }; // runtime contribution 25.86%