我最近刚接受采访的朋友听到了这个问题:
给定链表的头部,编写一个函数将头部与链表中的下一个元素交换,并将指针返回到新的头部。
例如:
i/p: 1->2,3,4,5 (the given head is 1)
o/p: 2->1,3,4,5
答案 0 :(得分:5)
假设
struct node {
struct node *next;
};
struct node *head;
然后解决方案看起来像
struct node *next = head->next;
if(next == NULL) return head; // nothing to swap
head->next = next->next;
next->next = head;
head = next;
return next;
答案 1 :(得分:1)
struct node* head;
struct node *tmp1,*tmp2;
tmp1=head; // save first node pointer
tmp2=head->next->next; // save third node pointer
head=head->next; // Move Head to the second node
head->next=tmp1; // swap
head->next->next=tmp2; // Restore the link to third node