66-Reorder List
- 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
}
};