• [LeetCode] Reorder List


    Given a singly linked list L: L0L1→…→Ln-1Ln, reorder it to: L0LnL1Ln-1L2Ln-2→…

    You must do this in-place without altering the nodes' values.

    For example, Given {1,2,3,4}, reorder it to {1,4,2,3}.

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        void reorderList(ListNode *head) {
            ListNode *p1=head,*p2=head,*pre;
            while(p2!=NULL && p2->next!=NULL)//找到后一半的开头节点head2;
            {
              pre = p1;
              p1 = p1->next;
              p2 = p2->next;
              p2 = p2->next;
            }
            
            if(p1 == p2)
                return;
            pre->next = NULL;//前一半的结尾标记
            ListNode *head2 = p1;
            
           stack<ListNode*> st;
           while(head2)
           {
             st.push(head2);
             head2 = head2->next;
           }
           p1 = head;
           ListNode *ph1,*ph2;
           while(p1 && !st.empty() )
           {
             ph1 = p1->next;
             ph2 = st.top();
             st.pop();
             p1->next = ph2;
             ph2->next = ph1;
             p1 = ph1;
           
           }
           if(!st.empty())
           {
             ph1 = st.top();
             ph1->next = NULL;
             st.pop();
             ph2->next = ph1;
           
           }
    
        }
    };

    (1)注意数据结构用stack比较方便,如果用遍历的方法,则会超时Time Limited Exceeded!
     (2)测试用例中,head包括NULL,1个结点、2个结点、3个结点、4个结点,把这几个测试正确,其余的应该就没有问题了。

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  • 原文地址:https://www.cnblogs.com/Xylophone/p/3842849.html
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