我被要求编写一个函数reverseHalves()来重新排列给定的链表,以便将节点的前半部分移到后半部分的后面。例如

给定链接列表[1 2 3 4 5 6],结果列表应为[4 5 6 1 2 3]
当给定的具有奇数个节点的链接列表,该列表应当是分裂与具有附加的节点中的第一半。也就是说,给定列表[1 2 3 4 5],结果列表应为[4 5 1 2 3]

但是我的函数将给出无限的输出...

typedef struct _listnode{
    int item;
    struct _listnode *next;
} ListNode;

typedef struct _linkedlist{
    int size;
    ListNode *head;
    ListNode *tail;
} LinkedList;


这是我使用的功能:

// printList will print out the value in every nodes until there is a NULL
void printList(LinkedList *ll);
ListNode * findNode(LinkedList *ll, int index);
int insertNode(LinkedList *ll, int index, int value);

void reverseHalves(LinkedList *ll)
{
    int index;
    ListNode *new_head, *new_tail;
    new_head = NULL;
    new_tail = NULL;

    // determine the index of new tail, and the new head which is index+1*
    index = (ll->size + 1) / 2;

    // get the new head by findNode func,whose index is index+1
    // make new_head point to the node found*
    new_head = findNode(ll, index);

    // make initial tail->next be the initial head*
    ll->tail->next = ll->head;

    // set the head to be the new head
    ll->head = new_head;

    insertNode(ll, ll->size, -1);
    new_tail = findNode(ll, ll->size);
    new_tail = NULL;
}

最佳答案

这是一个大致要点:

new_head_index = (ll->size+1) / 2;
new_tail_index = new_head_index - 1;

if (new_tail_index < 0)
    return;

new_head = findNode(ll,new_head_index);
new_tail = findNode(ll,new_tail_index);

ll->tail->next = ll->head;
new_tail->next = NULL
ll->head = new_head;

关于c - 倒排一半,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/15984552/

10-10 14:52