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_148.java
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_148.java
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package com.fishercoder.solutions;
import com.fishercoder.common.classes.ListNode;
public class _148 {
public static class Solution1 {
/**
* Credit: https://discuss.leetcode.com/topic/18100/java-merge-sort-solution
* But this is not using constant space.
*/
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
//Step 1: split the list into halves
ListNode prev = null;
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
prev = slow;
fast = fast.next.next;
slow = slow.next;
}
prev.next = null;
//step 2: sort each half
ListNode l1 = sortList(head);
ListNode l2 = sortList(slow);
//step 3: merge the two halves
return merge(l1, l2);
}
private ListNode merge(ListNode l1, ListNode l2) {
ListNode result = new ListNode(0);
ListNode tmp = result;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
tmp.next = l1;
l1 = l1.next;
} else {
tmp.next = l2;
l2 = l2.next;
}
tmp = tmp.next;
}
if (l1 != null) {
tmp.next = l1;
}
if (l2 != null) {
tmp.next = l2;
}
return result.next;
}
}
public static class Solution2 {
ListNode tail = new ListNode(0);
ListNode nextSubList = new ListNode(0);
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
int n = getCount(head);
ListNode start = head;
ListNode dummyHead = new ListNode(0);
for (int size = 1; size < n; size = size * 2) {
tail = dummyHead;
while (start != null) {
if (start.next == null) {
tail.next = start;
break;
}
ListNode mid = split(start, size);
merge(start, mid);
start = nextSubList;
}
start = dummyHead.next;
}
return dummyHead.next;
}
ListNode split(ListNode start, int size) {
ListNode midPrev = start;
ListNode end = start.next;
//use fast and slow approach to find middle and end of second linked list
for (int index = 1; index < size && (midPrev.next != null || end.next != null); index++) {
if (end.next != null) {
end = (end.next.next != null) ? end.next.next : end.next;
}
if (midPrev.next != null) {
midPrev = midPrev.next;
}
}
ListNode mid = midPrev.next;
midPrev.next = null;
nextSubList = end.next;
end.next = null;
// return the start of second linked list
return mid;
}
void merge(ListNode list1, ListNode list2) {
ListNode dummyHead = new ListNode(0);
ListNode newTail = dummyHead;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
newTail.next = list1;
list1 = list1.next;
newTail = newTail.next;
} else {
newTail.next = list2;
list2 = list2.next;
newTail = newTail.next;
}
}
newTail.next = (list1 != null) ? list1 : list2;
// traverse till the end of merged list to get the newTail
while (newTail.next != null) {
newTail = newTail.next;
}
// link the old tail with the head of merged list
tail.next = dummyHead.next;
// update the old tail to the new tail of merged list
tail = newTail;
}
int getCount(ListNode head) {
int cnt = 0;
ListNode ptr = head;
while (ptr != null) {
ptr = ptr.next;
cnt++;
}
return cnt;
}
}
public static class Solution3 {
/**
* Credit: https://leetcode.com/problems/sort-list/solution/ top down approach.
*/
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = getMid(head);
ListNode left = sortList(head);
ListNode right = sortList(mid);
return mergeList(left, right);
}
private ListNode mergeList(ListNode left, ListNode right) {
ListNode pre = new ListNode(-1);
ListNode tmp = pre;
while (left != null && right != null) {
if (left.val < right.val) {
tmp.next = left;
left = left.next;
} else {
tmp.next = right;
right = right.next;
}
tmp = tmp.next;
}
if (left != null) {
tmp.next = left;
} else if (right != null) {
tmp.next = right;
}
return pre.next;
}
private ListNode getMid(ListNode head) {
/**The key/trick is in this method:
* it directly uses this head to iterate, so that we could use this top down recursive approach.
* If we assign head to slow and fast pointers, then this algorithm will run into StackOverflow exception.
*
* This is an absolutely amazing method!*/
ListNode midPrev = null;
while (head != null && head.next != null) {
midPrev = (midPrev == null) ? head : midPrev.next;
head = head.next.next;
}
ListNode mid = midPrev.next;
midPrev.next = null;//this is the key, otherwise, StackOverflow exception will occur.
return mid;
}
}
}