劍指offer之鏈表

1.求鏈表倒數第K個節點

ahead指針先走K-1步，behind指針開始走，ahead走到末尾，此時behind剛好是倒數第K個

2.求鏈表中間節點

ahead指針一次走2步，behind一次走1步，當ahead走到末尾時，behind剛好是中間節點

3.判斷鏈表是否成環

ahead指針一次走2步，behind一次走1步，當ahead走到末尾時，還沒有追上behind說明無環。如果沒有到末尾就遇上了，說明有環。

4.鏈表反轉

2個指針相鄰同步後移,不停後者next改指向前者

ListNode* reverseList(ListNode* head)
{
    ListNode* result = NULL;//反轉後的頭節點
    ListNode* cur = head;
    ListNode* pre = NULL;

    while(cur != NULL)
    {
        ListNode* next = cur->next;//覆蓋前先記錄舊值
        if(next == NULL)
        {
            result = cur;//記錄尾節點
        }

        cur->next = pre;//改變cur節點next指針指向前者
        pre = cur;
        cur = next;//同步後移
    }
    return result;
}

5.鏈表合併

ListNode* mergeList(ListNode* head1, ListNode* head2)
{
    if(head1 == NULL)
        return head2;
    else if(head2 == NULL)
        return  head1;

    if(head1->data > head2->data)
    {//保證head1頭節點小於等於head2頭節點，合併後頭節點就是head1
        int temp = head2->data;
        head2->data = head1->data;
        head1->data = temp;
    }

    ListNode* cur1 = head1;
    while(head2 != NULL)//依次將鏈表2節點插入鏈表1合適位置
    {
        while(cur1->next != NULL && cur1->next->data < head2->data)
            cur1 = cur1->next;

        //此時cur1 <= head2 <= cur1->next或者cur1是尾節點但不大於head2,將head2插在cur後即可
         ListNode* tmp = head2;
        head2 = head2->next;

        tmp->next = cur1->next;
        cur1->next = tmp;
    }

    return head1;
}