LeetCode 148 - Sort List
https://gist.github.com/sing1ee/5949395
In QuickSortRecur(), we first call partition() which places pivot at correct position and returns pivot. After pivot is placed at correct position, we find tail node of left side (list before pivot) and recur for left list. Finally, we recur for right list.
http://algorithmsandme.in/2016/02/quick-sort-on-doubly-linked-list/
Read full article from QuickSort on Singly Linked List - GeeksforGeeks
https://gist.github.com/sing1ee/5949395
- 思路和数组的快速排序一样，都需要找到一个pivot元素、或者节点。然后将数组或者单向链表划分为两个部分，然后递归分别快排。
- 针对数组进行快排的时候，交换交换不同位置的数值，在分而治之完成之后，数据就是排序好的。那么单向链表是什么样的情况呢？除了交换节点值之外，是否有其他更好的方法呢？可以修改指针，不进行数值交换。这可以获取更高的效率。
- 在修改指针的过程中，会产生新的头指针以及尾指针，要记录下来。在partition之后，要将小于pivot的的部分、pivot、以及大于pivot的部分重新串起来成为一个singly linked list。
public ListNode quickSortRecur(ListNode head, ListNode tail) { if (head == tail) return head; ListNode pivot = tail, cur = head, pre = null, newhead = null; while(pivot != cur) { if(cur.val < pivot.val) { if (null == newhead) newhead = cur; pre = cur; cur = cur.next; } else { tail.next = cur; tail = tail.next; ListNode tmp = cur.next; cur.next = null; cur = tmp; if (null != pre) { pre.next = cur; } } } if (pre != null) { newhead = quickSortRecur(newhead, pre); ListNode tmp = newhead; while(pivot != tmp.next) tmp = tmp.next; tmp.next = pivot; }else { newhead = pivot; } if (pivot != tail) pivot.next = quickSortRecur(pivot.next, tail); return newhead; }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://github.com/AlexYursha/problemsolving/blob/master/sort/QuickSort.java
QuickSort on Singly Linked List - GeeksforGeeks
In partition(), we consider last element as pivot. We traverse through the current list and if a node has value greater than pivot, we move it after tail. If the node has smaller value, we keep it at its current position.
private static <T extends Comparable<T>> LinkedList<T> quickSortLinkedList(LinkedList<T> list) {
if (list == null) {
throw new NullPointerException();
}
if (list.size() < 2) {
return list;
}
LinkedList<T> left = new LinkedList<T>();
T pivot = list.removeFirst();
Iterator<T> iterator = list.listIterator(0);
while (iterator.hasNext()) {
T item = iterator.next();
if (pivot.compareTo(item) > 0) {
left.add(item);
iterator.remove();
}
}
iterator = null;
left = quickSortLinkedList(left);
list = quickSortLinkedList(list);
left.add(pivot);
left.addAll(list);
return left;
}
QuickSort on Singly Linked List - GeeksforGeeks
In QuickSortRecur(), we first call partition() which places pivot at correct position and returns pivot. After pivot is placed at correct position, we find tail node of left side (list before pivot) and recur for left list. Finally, we recur for right list.
struct node *getTail(struct node *cur){ while (cur != NULL && cur->next != NULL) cur = cur->next; return cur;}// Partitions the list taking the last element as the pivotstruct node *partition(struct node *head, struct node *end, struct node **newHead, struct node **newEnd){ struct node *pivot = end; struct node *prev = NULL, *cur = head, *tail = pivot; // During partition, both the head and end of the list might change // which is updated in the newHead and newEnd variables while (cur != pivot) { if (cur->data < pivot->data) { // First node that has a value less than the pivot - becomes // the new head if ((*newHead) == NULL) (*newHead) = cur; prev = cur; cur = cur->next; } else // If cur node is greater than pivot { // Move cur node to next of tail, and change tail if (prev) prev->next = cur->next; struct node *tmp = cur->next; cur->next = NULL; tail->next = cur; tail = cur; cur = tmp; } } // If the pivot data is the smallest element in the current list, // pivot becomes the head if ((*newHead) == NULL) (*newHead) = pivot; // Update newEnd to the current last node (*newEnd) = tail; // Return the pivot node return pivot;}//here the sorting happens exclusive of the end nodestruct node *quickSortRecur(struct node *head, struct node *end){ // base condition if (!head || head == end) return head; node *newHead = NULL, *newEnd = NULL; // Partition the list, newHead and newEnd will be updated // by the partition function struct node *pivot = partition(head, end, &newHead, &newEnd); // If pivot is the smallest element - no need to recur for // the left part. if (newHead != pivot) { // Set the node before the pivot node as NULL struct node *tmp = newHead; while (tmp->next != pivot) tmp = tmp->next; tmp->next = NULL; // Recur for the list before pivot newHead = quickSortRecur(newHead, tmp); // Change next of last node of the left half to pivot tmp = getTail(newHead); tmp->next = pivot; } // Recur for the list after the pivot element pivot->next = quickSortRecur(pivot->next, newEnd); return newHead;}// The main function for quick sort. This is a wrapper over recursive// function quickSortRecur()void quickSort(struct node **headRef){ (*headRef) = quickSortRecur(*headRef, getTail(*headRef)); return;}Read full article from QuickSort on Singly Linked List - GeeksforGeeks
