Swap Nodes in Pairs
题目描述:
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.
这题不难,直接贴两个我中意的代码:
solution1:
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode* swapPairs(ListNode *head) { if (head == NULL || head->next == NULL) return head; ListNode *list = head->next; head->next = swapPairs(list->next); list->next = head; return list; }
参考链接:https://leetcode.com/discuss/16246/my-simple-recursive-solution
solution2:
ListNode* swap(ListNode *next1, ListNode *next2) { next1->next = next2->next; next2->next = next1; return next2; } ListNode* swapPairs(ListNode *head) { if (head == NULL || head->next == NULL) return head; ListNode *start = new ListNode(0); start->next = head; for (ListNode *cur = start;cur->next && cur->next->next;cur = cur->next->next) { cur->next = swap(cur->next, cur->next->next); } return start->next; }
ps:
递归解法真的很赞,有木有。