Sort a linked list in O(n log n) time using constant space complexity.
Have you met this question in a real interview?
Yes
Example
Given 1->3->2->null, sort it to 1->2->3->null.
Challenge
Solve it by merge sort & quick sort separately.
LeetCode上的原题，请参见我之前的博客Sort List。
解法一：
class Solution {
public:
/**
* @param head: The first node of linked list.
* @return: You should return the head of the sorted linked list,
using constant space complexity.
*/
ListNode *sortList(ListNode *head) {
if (!head || !head->next) return head;
ListNode *fast = head, *slow = head, *pre = head;
while (fast && fast->next) {
pre = slow;
slow = slow->next;
fast = fast->next->next;
}
pre->next = NULL;
return merge(sortList(head), sortList(slow));
}
ListNode *merge(ListNode *l1, ListNode *l2) {
if (!l1) return l2;
if (!l2) return l1;
if (l1->val < l2->val) {
l1->next = merge(l1->next, l2);
return l1;
} else {
l2->next = merge(l1, l2->next);
return l2;
}
}
};
解法二：
class Solution {
public:
/**
* @param head: The first node of linked list.
* @return: You should return the head of the sorted linked list,
using constant space complexity.
*/
ListNode *sortList(ListNode *head) {
if (!head || !head->next) return head;
ListNode *fast = head, *slow = head, *pre = head;
while (fast && fast->next) {
pre = slow;
slow = slow->next;
fast = fast->next->next;
}
pre->next = NULL;
return merge(sortList(head), sortList(slow));
}
ListNode *merge(ListNode *l1, ListNode *l2) {
ListNode *dummy = new ListNode(-);
ListNode *cur = dummy;
while (l1 && l2) {
if (l1->val < l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
if (l1) cur->next = l1;
if (l2) cur->next = l2;
return dummy->next;
}
};