- Sort List
Sort a linked list in O(n log n) time using constant space complexity.
Example 1:
Input: 4->2->1->3
Output: 1->2->3->4
Example 2:
Input: -1->5->3->4->0
Output: -1->0->3->4->5
解法1 归并排序。用两个函数实现:
- merge:将两个有序链表合在一起
- merge_sort:将无序链表排序
找中间位置用双指针
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* sortList(ListNode* head) {
if(head == NULL || head->next == NULL)return head;
ListNode *mid, *pre;
find_mid(head, mid, pre);
pre->next = NULL;
return merge(sortList(head), sortList(mid));
}
// 找中点时,需要对mid和pre做修改,因此需要传引用
void find_mid(ListNode* s, ListNode* &mid, ListNode* &pre){
ListNode *pp = s;
mid = s;
while(pp && pp->next){
pre = mid;
mid = mid->next;
pp = pp->next->next;
}
}
ListNode* merge(ListNode* s1, ListNode* s2){
ListNode *head = new ListNode, *p = s1, *q = s2;
ListNode *cur = head;
while(p != NULL && q != NULL){
if(p->val < q->val){
cur->next = p;
p = p->next;
}else{
cur->next = q;
q = q->next;
}
cur = cur->next;
}
if(p != NULL)cur->next = p;
else cur->next = q;
return head->next;
}
};
解法2 快速排序。partition过程中,可以用两个链表small和large分别存储pivot左侧和右侧的数据
class Solution {
public:
ListNode* sortList(ListNode* head) {
if(!head || !head->next) return head;
ListNode* cur = head->next;
ListNode* small = new ListNode(0);
ListNode* large = new ListNode(0);
ListNode* sp = small;
ListNode* lp = large;
// partition
while(cur){
if(cur->val<head->val){
sp->next = cur;
sp = cur;
}
else{
lp->next = cur;
lp = cur;
}
cur = cur->next;
}
sp->next = NULL;
lp->next = NULL;
small=sortList(small->next);
large=sortList(large->next);
cur = small;
if(cur){
while(cur->next) cur = cur->next;
cur->next = head;
head->next = large;
return small;
}else{
head->next = large;
return head;
}
}
};