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148. Sort List
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148. Sort List
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/*
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
*/
/*
Idea: Merge Sort for linked list.
Linked list can be devided into 2 lists not exactly from the middle.
devide into 2 lists
4 2 3 1
4 3 2 1
4 3 2 1
merge sorted lists
2 4 1 3
1 2 3 4
*/
class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode a = new ListNode(head.val);
ListNode aHead = a;
ListNode b = new ListNode(head.next.val);
ListNode bHead = b;
head = head.next.next;
divideLinkedListInto2Lists(head, a, b);
a = sortList(aHead);
b = sortList(bHead);
return mergeSortedLinkedLists(a, b);
}
private void divideLinkedListInto2Lists(ListNode head, ListNode a, ListNode b) {
if (head == null) {
return;
} else {
while (head != null) {
a.next = new ListNode(head.val);
a = a.next;
head = head.next;
if (head != null) {
b.next = new ListNode(head.val);
b = b.next;
head = head.next;
}
}
}
}
private ListNode mergeSortedLinkedLists(ListNode a, ListNode b) {
if (a == null) return b;
if (b == null) return a;
ListNode merged;
if (a.val < b.val) {
merged = a;
a = a.next;
} else {
merged = b;
b = b.next;
}
ListNode headRef = merged;
while (a != null || b != null) {
if (a != null && b != null) {
if (a.val < b.val) {
merged.next = a;
a = a.next;
} else {
merged.next = b;
b = b.next;
}
} else if (a != null) {
merged.next = a;
a = a.next;
} else if (b != null) {
merged.next = b;
b = b.next;
} else {
return headRef;
}
merged = merged.next;
}
return headRef;
}
}
/*
Bad solution N^2
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
boolean needToBeSorted = false;
ListNode curr = head;
while (curr != null) {
ListNode next = curr.next;
if (next == null) {
if (needToBeSorted) {
return sortList(head);
} else {
return head;
}
}
if (curr.val > next.val) {
needToBeSorted = true;
swapValues(curr, curr.next);
}
curr = curr.next;
}
if (needToBeSorted) {
return sortList(head);
} else {
return head;
}
}
private void swapValues(ListNode a, ListNode b) {
a.val = a.val + b.val;
b.val = a.val - b.val;
a.val = a.val - b.val;
}
*/