Sort a linked list in O(n log n) time using constant space complexity.
X. Merge Sort
http://www.cnblogs.com/springfor/p/3869372.html
https://discuss.leetcode.com/topic/18100/java-merge-sort-solution
http://www.jiuzhang.com/solutions/sort-list/
X. Quick sort
https://www.fyears.org/2016/05/quicksort-in-array-and-linked-list.html
http://www.jiuzhang.com/solutions/sort-list/
https://discuss.leetcode.com/topic/5055/a-minor-tweak-on-quick-sort-to-solve-the-time-limit-exceeded-issue
QuickSort on Singly Linked List - GeeksforGeeks
http://rleetcode.blogspot.com/2014/01/sort-list-java.html
http://www.geeksforgeeks.org/quicksort-for-linked-list/
http://www.geeksforgeeks.org/why-quick-sort-preferred-for-arrays-and-merge-sort-for-linked-lists/
X. Merge Sort
http://www.cnblogs.com/springfor/p/3869372.html
https://discuss.leetcode.com/topic/18100/java-merge-sort-solution
不过这道题要找linkedlist中点，那当然就要用最经典的faster和slower方法，faster速度是slower的两倍，当faster到链尾时，slower就是中点，slower的next是下一半的开始点。
我们可以使用divide and conquer的方式，依次递归的对链表左右两半进行排序就可以了
1 public ListNode sortList(ListNode head) {
2 if(head == null|| head.next == null)
3 return head;
4 ListNode slow = head, fast = head, firsthalf = head;
5 while(fast.next!=null&&fast.next.next!=null){
6 slow = slow.next;
7 fast = fast.next.next;
8 }
9 ListNode secondhalf = slow.next;
10 slow.next = null;
11
12 ListNode leftlist = null, rightlist =null;
13 if(firsthalf!=secondhalf){
14 leftlist = sortList(firsthalf);
15 rightlist = sortList(secondhalf);
16 }
17 return mergeTwoLists(leftlist, rightlist);
18 }
19
20 public ListNode mergeTwoLists(ListNode leftlist, ListNode rightlist){
21 if(rightlist == null)
22 return leftlist;
23 if(leftlist == null)
24 return rightlist;
25
26 ListNode fakehead = new ListNode(-1);
27 ListNode ptr = fakehead;
28 while(rightlist!=null&&leftlist!=null){
29 if(rightlist.val<leftlist.val){
30 ptr.next = rightlist;
31 ptr = ptr.next;
32 rightlist = rightlist.next;
33 }else{
34 ptr.next = leftlist;
35 ptr = ptr.next;
36 leftlist = leftlist.next;
37 }
38 }
39
40 if(rightlist!=null)
41 ptr.next = rightlist;
42 if(leftlist!=null)
43 ptr.next = leftlist;
44
45 return fakehead.next;
46 }
http://www.dontforgettothink.com/2011/11/23/merge-sort-of-linked-list/2 if(head == null|| head.next == null)
3 return head;
4 ListNode slow = head, fast = head, firsthalf = head;
5 while(fast.next!=null&&fast.next.next!=null){
6 slow = slow.next;
7 fast = fast.next.next;
8 }
9 ListNode secondhalf = slow.next;
10 slow.next = null;
11
12 ListNode leftlist = null, rightlist =null;
13 if(firsthalf!=secondhalf){
14 leftlist = sortList(firsthalf);
15 rightlist = sortList(secondhalf);
16 }
17 return mergeTwoLists(leftlist, rightlist);
18 }
19
20 public ListNode mergeTwoLists(ListNode leftlist, ListNode rightlist){
21 if(rightlist == null)
22 return leftlist;
23 if(leftlist == null)
24 return rightlist;
25
26 ListNode fakehead = new ListNode(-1);
27 ListNode ptr = fakehead;
28 while(rightlist!=null&&leftlist!=null){
29 if(rightlist.val<leftlist.val){
30 ptr.next = rightlist;
31 ptr = ptr.next;
32 rightlist = rightlist.next;
33 }else{
34 ptr.next = leftlist;
35 ptr = ptr.next;
36 leftlist = leftlist.next;
37 }
38 }
39
40 if(rightlist!=null)
41 ptr.next = rightlist;
42 if(leftlist!=null)
43 ptr.next = leftlist;
44
45 return fakehead.next;
46 }
http://www.jiuzhang.com/solutions/sort-list/
private ListNode findMiddle(ListNode head) { ListNode slow = head, fast = head.next; while (fast != null && fast.next != null) { fast = fast.next.next; slow = slow.next; } return slow; } private ListNode merge(ListNode head1, ListNode head2) { ListNode dummy = new ListNode(0); ListNode tail = dummy; while (head1 != null && head2 != null) { if (head1.val < head2.val) { tail.next = head1; head1 = head1.next; } else { tail.next = head2; head2 = head2.next; } tail = tail.next; } if (head1 != null) { tail.next = head1; } else { tail.next = head2; } return dummy.next; } public ListNode sortList(ListNode head) { if (head == null || head.next == null) { return head; } ListNode mid = findMiddle(head); ListNode right = sortList(mid.next); mid.next = null; ListNode left = sortList(head); return merge(left, right); }http://www.cnblogs.com/grandyang/p/4249905.html
public ListNode sortList(ListNode head) { if (head == null || head.next == null) return head; ListNode slow = head, fast = head, pre = head; while (fast != null && fast.next != null) { pre = slow; slow = slow.next; fast = fast.next.next; } pre.next = null; return merge(sortList(head), sortList(slow)); } public ListNode merge(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(-1); ListNode cur = dummy; while (l1 != null && l2 != null) { if (l1.val < l2.val) { cur.next = l1; l1 = l1.next; } else { cur.next = l2; l2 = l2.next; } cur = cur.next; } if (l1 != null) cur.next = l1; if (l2 != null) cur.next = l2; return dummy.next; }
X. Quick sort
https://www.fyears.org/2016/05/quicksort-in-array-and-linked-list.html
public ListNode quickSort(ListNode head) {
if (head == null || head.next == null){
return head;
}
return quick(head, null);
}
private ListNode quick(ListNode start, ListNode end){
if (start == null || start == end || start.next == end){
return start;
}
ListNode[] result = partition(start, end);
ListNode resultLeft = quick(result[0], result[1]);
ListNode resultRight = quick(result[1].next, end);
return resultLeft;
}
private ListNode[] partition(ListNode start, ListNode end){
// start inclusive
// end exclusive
// return the new start node and the pivot node
if (start == null || start == end || start.next == end){
return new ListNode[] {start, start};
}
ListNode dummy = new ListNode(0);
dummy.next = start;
for (ListNode j = start; j != null && j.next != null && j.next != end; j = j.next) {
while (j.next != null && j.next.value <= start.value){
ListNode tmp = j.next;
j.next = j.next.next;
tmp.next = dummy.next;
dummy.next = tmp;
}
}
return new ListNode[] {dummy.next, start};
}
https://discuss.leetcode.com/topic/5055/a-minor-tweak-on-quick-sort-to-solve-the-time-limit-exceeded-issue
public ListNode sortList(ListNode head) {
if(head == null || head.next == null){
return head;
}
ListNode pivot = head;
head = head.next;
ListNode bigHead = null, smallHead=null;
ListNode bigIter = null, smallIter=null;
ListNode pivotIter = pivot;
while(head != null){
ListNode iter = head.next;
if(head.val > pivot.val){
if(bigHead == null){
bigHead = head;
bigIter = bigHead;
}else{
bigIter.next = head;
bigIter = bigIter.next;
}
}else if(head.val == pivot.val){
pivotIter.next = head;
pivotIter = pivotIter.next;
}else{
if(smallHead == null){
smallHead = head;
smallIter = smallHead;
}else{
smallIter.next = head;
smallIter = smallIter.next;
}
}
head = iter;
}
// mark the ends for the big/small sub lists.
if(smallIter != null){
smallIter.next = null;
}
pivotIter.next = null;
if(bigIter != null){
bigIter.next = null;
}
ListNode sortBig = sortList(bigHead);
ListNode sortSmall = sortList(smallHead);
if(sortSmall == null){
pivotIter.next = sortBig;
return pivot;
} else {
smallIter = sortSmall;
while (smallIter.next != null) {
smallIter = smallIter.next;
}
// reach the tail of the small sublist
smallIter.next = pivot;
// concatenate the two sublists.
pivotIter.next = sortBig;
return sortSmall;
}
}public ListNode sortList(ListNode head) {
if(head == null || head.next == null) return head;
ListNode left = new ListNode(0), leftHead = left;
ListNode right = new ListNode(0), rightHead = right;
ListNode mid = new ListNode(0), midHead = mid;
int val = head.val;
while(head != null){
if(head.val < val){
left.next = head;
left = head;
}
if(head.val > val){
right.next = head;
right = head;
}
if(head.val == val){
mid.next = head;
mid = head;
}
head = head.next;
}
left.next = null;
right.next = null;
mid.next = null;
return concat(sortList(leftHead.next),midHead.next,sortList(rightHead.next));
}
public ListNode concat(ListNode left, ListNode mid, ListNode right){
ListNode LeftTail = getTail(left);
ListNode midTail = getTail(mid);
midTail.next = right;
if(LeftTail != null) {
LeftTail.next = mid;
return left;
} else {
return mid;
}
}
public ListNode getTail(ListNode head){
if(head == null) return head;
while(head.next != null) head = head.next;
return head;
}ListNode quickSort(final ListNode h){
if(h == null || h.next == null)
return h;
/*split into three list*/
ListNode fakesmall = new ListNode(0), small = fakesmall;
ListNode fakelarge = new ListNode(0), large = fakelarge;
ListNode fakeequal = new ListNode(0), equal = fakeequal;
ListNode cur = h; // pivot is h.
while(cur != null){
if(cur.val < h.val){
small.next = cur;
small = small.next;
}
else if(cur.val == h.val){
equal.next = cur;
equal = equal.next;
}
else{
large.next = cur;
large = large.next;
}
cur = cur.next;
}
// put an end.
small.next = equal.next = large.next = null;
// merge them and return . merge reusing below one. merge for quicksort should be simplified.
return merge(merge(quickSort(fakesmall.next), quickSort(fakelarge.next)),fakeequal.next) ;
}
/*mrege sort*/
ListNode mergeSort(ListNode h){
if(h == null || h.next == null)
return h;
/*find cutting point*/
ListNode slow = h, cut = null, fast = h;
while(fast != null && fast.next != null){
cut = slow;
slow = slow.next;
fast = fast.next.next;
}
/*cut*/
cut.next = null;
return merge(mergeSort(h),mergeSort(slow));
}
ListNode merge(ListNode h, ListNode m){
ListNode fake = new ListNode(0), cur = fake;
while(h != null && m != null){
if(h.val < m.val){
cur.next = h;
h = h.next;
}
else{
cur.next = m;
m = m.next;
}
cur = cur.next;
}
cur.next = (h == null ? m : h);
return fake.next;
}QuickSort on Singly Linked List - GeeksforGeeks
public static ListNode sort(ListNode head){
if (head == null)
return head;
ListNode pivot = new ListNode(head.val);
head = head.next; // remove pivot
ListNode lessDummy = new ListNode(0);
ListNode moreDummy = new ListNode(0);
ListNode lessP = lessDummy;
ListNode moreP = moreDummy;
ListNode p = head;
while (p != null){
if (p.val < pivot.val){
lessP.next = p;
lessP = lessP.next;
p = p.next;
}else {
moreP.next = p;
moreP = moreP.next;
p = p.next;
}
}
lessP.next=null;//<-----当年少了这两个
moreP.next= null;//<----
lessDummy = sort(lessDummy.next);
moreDummy = sort(moreDummy.next);
lessP = lessDummy;
moreP = moreDummy;
if(lessP != null) {
while (lessP.next != null) {
lessP = lessP.next;
}
lessP.next = pivot;
pivot.next = moreP;
return lessDummy;
}
else {
pivot.next = moreDummy;
return pivot;
}
}
http://rleetcode.blogspot.com/2014/01/sort-list-java.html
http://www.geeksforgeeks.org/quicksort-for-linked-list/
/* a node of the doubly linked list */ static class Node{ private int data; private Node next; private Node prev; Node(int d){ data = d; next = null; prev = null; } } // A utility function to find last node of linked list Node lastNode(Node node){ while(node.next!=null) node = node.next; return node; } /* Considers last element as pivot, places the pivot element at its correct position in sorted array, and places all smaller (smaller than pivot) to left of pivot and all greater elements to right of pivot */ Node partition(Node l,Node h) { // set pivot as h element int x = h.data; // similar to i = l-1 for array implementation Node i = l.prev; // Similar to "for (int j = l; j <= h- 1; j++)" for(Node j=l; j!=h; j=j.next) { if(j.data <= x) { // Similar to i++ for array i = (i==null) ? l : i.next; int temp = i.data; i.data = j.data; j.data = temp; } } i = (i==null) ? l : i.next; // Similar to i++ int temp = i.data; i.data = h.data; h.data = temp; return i; } /* A recursive implementation of quicksort for linked list */ void _quickSort(Node l,Node h) { if(h!=null && l!=h && l!=h.next){ Node temp = partition(l,h); _quickSort(l,temp.prev); _quickSort(temp.next,h); } } // The main function to sort a linked list. It mainly calls _quickSort() public void quickSort(Node node) { // Find last node Node head = lastNode(node); // Call the recursive QuickSort _quickSort(node,head); }int partition (int arr[], int l, int h){ int x = arr[h]; int i = (l - 1); for (int j = l; j <= h- 1; j++) { if (arr[j] <= x) { i++; swap (&arr[i], &arr[j]); } } swap (&arr[i + 1], &arr[h]); return (i + 1);}/* A[] --> Array to be sorted, l --> Starting index, h --> Ending index */void quickSort(int A[], int l, int h){ if (l < h) { int p = partition(A, l, h); /* Partitioning index */ quickSort(A, l, p - 1); quickSort(A, p + 1, h); }}
Quick Sort in its general form is an in-place sort (i.e. it doesn’t require any extra storage) whereas merge sort requires O(N) extra storage, N denoting the array size which may be quite expensive. Allocating and de-allocating the extra space used for merge sort increases the running time of the algorithm. Comparing average complexity we find that both type of sorts have O(NlogN) average complexity but the constants differ. For arrays, merge sort loses due to the use of extra O(N) storage space.
Quick Sort is also a cache friendly sorting algorithm as it has good locality of reference when used for arrays.
Quick Sort is also tail recursive, therefore tail call optimizations is done.
In case of linked lists the case is different mainly due to difference in memory allocation of arrays and linked lists. Unlike arrays, linked list nodes may not be adjacent in memory. Unlike array, in linked list, we can insert items in the middle in O(1) extra space and O(1) time. Therefore merge operation of merge sort can be implemented without extra space for linked lists.
In arrays, we can do random access as elements are continuous in memory. Let us say we have an integer (4-byte) array A and let the address of A[0] be x then to access A[i], we can directly access the memory at (x + i*4). Unlike arrays, we can not do random access in linked list. Quick Sort requires a lot of this kind of access. In linked list to access i’th index, we have to travel each and every node from the head to i’th node as we don’t have continuous block of memory. Therefore, the overhead increases for quick sort. Merge sort accesses data sequentially and the need of random access is low.
