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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}
.
Linked List
分析:用stack保存link List,然后从后向前弹出,
cur < total/2 时完成,最后把最后一个节点的next设置成NULL
/** * 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) { if(head == NULL) return ; ListNode* pFront = head; stack<ListNode*> st; int totalNum = 0; int curNum = 0; while(pFront) { st.push(pFront); pFront= pFront->next; totalNum++; } pFront = head; ListNode* pRear = NULL; ListNode* tmp = NULL; while(curNum < totalNum/2) { pRear = st.top(); st.pop(); tmp = pFront->next; pFront->next = pRear; pRear->next = tmp; pFront= tmp; curNum ++; } //mark the end pFront->next = NULL; } };
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原文地址:http://www.cnblogs.com/diegodu/p/4597791.html