Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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。