66-Reorder List

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Reorder List My Submissions QuestionEditorial Solution
Total Accepted: 64392 Total Submissions: 281830 Difficulty: Medium
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}.

思路：
1.找到中点，前半部分最后一个点
a.找到后半部分第一个点
b.后半部分入栈
c.遍历前半部分，间隔性依次链接栈中节点

时间复杂度：
空间复杂度：

/** * 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) {        int len=0;        if(head==NULL||head->next==NULL)return;        if(head->next->next==NULL)return;        ListNode *p=head,*fir_end,*sec_beg;        while(p){            len++;            p = p->next;        }        int mid=(len%2==0)?len/2:len/2+1;        mid = mid -1;        fir_end = head;        while(mid--){            fir_end=fir_end->next;        }        sec_beg = fir_end->next;        stack<ListNode*> slist;        while(sec_beg!=NULL){            slist.push(sec_beg);            sec_beg = sec_beg->next;        }        while(!slist.empty()){  //如果栈中有元素未被链接起来            ListNode *tmp=head->next,*stop=slist.top(); //保存前半部分当前节点下一个节点            head->next = stop;   //链接栈中元素            stop->next = tmp;    //栈中元素链接到原来当前节点的下一元素，相当于在中间插入            slist.pop();             head = tmp;        }        if(head!=NULL)head->next = NULL;//如果链长为奇数，最后一个元素指向空    }};

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