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题目: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.
解决思路:初始状态下,让指针p指向第一个节点,q指向第二个节点,当q不为空时,交换两节点的值,再让p指向q后面的节点,q指向p后面的节点,交换节点值,一次循环管直到遇到空节点。代码如下:
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 *p = head; ListNode *q = p->next; while (q) { int temp; temp = p->val; p->val = q->val; q->val = temp; if (q->next != NULL) { p = q->next; if (p->next != NULL) q = p->next; else break; } else break; } return head; }
【LeetCode OJ】Swap Nodes in Pairs
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原文地址:http://blog.csdn.net/xujian_2014/article/details/44827685