题目描述
对一个单链表原地(in-place)排序。即直接对链表结点排序。返回排序后链表的头结点。
链表结点的定义为(请不要在代码中再次定义该结构):
C/C++struct ListNode {
int val;
ListNode *next;
}
渣做法!N^2
ListNode* insert(ListNode* head, ListNode* item) {
if (item == NULL) return head;
if (head == NULL) { item->next = NULL; return item; }
if (item->val < head->val) { item->next = head; return item;}
ListNode* prev = head, *cur = head->next;
while (cur != NULL) {
if (cur->val < item->val) prev = cur, cur = cur->next;
else {
prev->next = item;
item->next = cur;
return head;
}
}
if (cur == NULL) {
prev->next = item;
item->next = NULL;
}
return head;
}
ListNode* sortLinkList(ListNode *head) {
if (!head) return head;
ListNode *cur = head, *newhead = NULL, *next;
while (cur != NULL) {
next = cur->next;
newhead = insert(newhead, cur);
cur = next;
}
return newhead;
}
merge sort
ListNode* merge(ListNode* a, ListNode* b) {
if (a == NULL || b == NULL) return a == NULL ? b : a;
ListNode* head, *cur;
if (a->val < b->val) head = a, a = a->next;
else head = b, b = b->next;
head->next = NULL;
cur = head;
while (a != NULL && b != NULL) {
if (a->val < b->val) cur->next = a, cur = cur->next, a = a->next;
else cur->next = b, cur = cur->next, b = b->next;
}
if (a != NULL) cur->next = a;
else if (b != NULL) cur->next = b;
else cur->next = NULL;
return head;
}
ListNode* mergeSort(ListNode* start, int len) {
if (len <= 0) return NULL;
if (len == 1) return start;
int mid = len / 2, t;
ListNode *midnode = start;
t = mid;
while (t-- > 1) midnode = midnode->next;
ListNode *start2 = midnode->next;
midnode->next = NULL;
return merge(mergeSort(start, mid), mergeSort(start2, len - mid));
}
ListNode* sortLinkList(ListNode *head) {
ListNode* cur = head;
int n = 0;
while (cur != NULL) n++, cur = cur->next;
return mergeSort(head, n);
}
本文介绍了一种单链表的原地排序方法,包括效率较低的插入排序实现及更高效的归并排序实现。通过对链表进行原地操作,避免了额外的空间开销。
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