标签：each 中间 ++ 要求 code exit and one length
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：归并排序。由于有时间和空间复杂度的要求。把链表从中间分开，递归下去，都最后两个node时开始合并，返回上一层继续合并，直到结束。找中间点的方法可以用快慢指针，快指针走2步，慢指针走1步。也可以求出链表长度，再分开链表。
Java:
public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null)
return head;
// step 1. cut the list to two halves
ListNode prev = null, slow = head, fast = head;
while (fast != null && fast.next != null) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
prev.next = null;
// step 2. sort each half
ListNode l1 = sortList(head);
ListNode l2 = sortList(slow);
// step 3. merge l1 and l2
return merge(l1, l2);
}
ListNode merge(ListNode l1, ListNode l2) {
ListNode l = new ListNode(0), p = l;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
p.next = l1;
l1 = l1.next;
} else {
p.next = l2;
l2 = l2.next;
}
p = p.next;
}
if (l1 != null)
p.next = l1;
if (l2 != null)
p.next = l2;
return l.next;
}
}
Python:
class Solution(object):
def merge(self, h1, h2):
dummy = tail = ListNode(None)
while h1 and h2:
if h1.val < h2.val:
tail.next, tail, h1 = h1, h1, h1.next
else:
tail.next, tail, h2 = h2, h2, h2.next
tail.next = h1 or h2
return dummy.next
def sortList(self, head):
if not head or not head.next:
return head
pre, slow, fast = None, head, head
while fast and fast.next:
pre, slow, fast = slow, slow.next, fast.next.next
pre.next = None
return self.merge(*map(self.sortList, (head, slow)))
C++:
class Solution {
public:
ListNode *sortList(ListNode *head) {
if(!head || !(head->next)) return head;
//get the linked list‘s length
ListNode* cur = head;
int length = 0;
while(cur){
length++;
cur = cur->next;
}
ListNode dummy(0);
dummy.next = head;
ListNode *left, *right, *tail;
for(int step = 1; step < length; step <<= 1){
cur = dummy.next;
tail = &dummy;
while(cur){
left = cur;
right = split(left, step);
cur = split(right,step);
tail = merge(left, right, tail);
}
}
return dummy.next;
}
private:
/**
* Divide the linked list into two lists,
* while the first list contains first n ndoes
* return the second list‘s head
*/
ListNode* split(ListNode *head, int n){
//if(!head) return NULL;
for(int i = 1; head && i < n; i++) head = head->next;
if(!head) return NULL;
ListNode *second = head->next;
head->next = NULL;
return second;
}
/**
* merge the two sorted linked list l1 and l2,
* then append the merged sorted linked list to the node head
* return the tail of the merged sorted linked list
*/
ListNode* merge(ListNode* l1, ListNode* l2, ListNode* head){
ListNode *cur = head;
while(l1 && l2){
if(l1->val > l2->val){
cur->next = l2;
cur = l2;
l2 = l2->next;
}
else{
cur->next = l1;
cur = l1;
l1 = l1->next;
}
}
cur->next = (l1 ? l1 : l2);
while(cur->next) cur = cur->next;
return cur;
}
};
[LeetCode] 148. Sort List 链表排序
标签：each 中间 ++ 要求 code exit and one length
原文地址：https://www.cnblogs.com/lightwindy/p/9609026.html
