刷题——SortList

/* * Sort a linked list in O(n log n) time using constant space complexity. * 本题综合利用快慢指针与归并排序，收藏了。 */

public class SortList {//快慢指针找中间节点public ListNode findCenterNode(ListNode head){if (head==null || head.next==null) {return head;}ListNode slowNode = head;ListNode fastNode = head;while (fastNode.next!=null && fastNode.next.next!=null) {slowNode = slowNode.next;fastNode = fastNode.next.next;}return slowNode;}//合并两个链表public ListNode merge(ListNode left,ListNode right){if (left==null || right==null) {return left==null?right:left;}ListNode head = new ListNode(-1);ListNode tail = head;while (left!=null && right!=null) {if (left.val<=right.val) {tail.next = left;left = left.next;}else {tail.next = right;right = right.next;}tail = tail.next;}tail.next = (left!=null)?left:right;return head.next;}public ListNode sortList(ListNode head) {        if (head==null || head.next==null) return head;ListNode centerNode = findCenterNode(head);ListNode nextNode = centerNode.next;centerNode.next = null;ListNode leftNode = sortList(head);ListNode rightNode = sortList(nextNode);return merge(leftNode,rightNode);    }public static void main(String[] args) {// TODO Auto-generated method stubListNode node0 = new ListNode(5);ListNode node1 = new ListNode(4);ListNode node2 = new ListNode(1);ListNode node3 = new ListNode(7);ListNode node4 = new ListNode(6);node0.next = node1;node1.next = node2;node2.next = node3;node3.next = node4;ListNode head = new SortList().sortList(node0);while (head!=null) {System.out.print(head.val+" ");head=head.next;}System.out.println();}}