Sort List 链表排序@LeetCode

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要求时间为O(nlogn)，最适合就是merge sort，每次从n/2处断开，对两段递归sort，然后再merge起来。

空间是O(logn)用了栈空间

package Level4;import Utility.ListNode;/** * Sort List *  * Sort a linked list in O(n log n) time using constant space complexity. * */public class S144 {public static void main(String[] args) {int[] list = {1,4,2,3};ListNode head = ListNode.create(list);ListNode h = sortList(head);h.print();}public static ListNode sortList(ListNode head) {if(head==null || head.next==null){// 链表没有元素或是只有一个元素的情况直接返回return head;}ListNode fast = head;ListNode slow = head;ListNode preSlow = head;// 找到中间节点的前一个while(fast!=null && fast.next!=null){fast = fast.next.next;preSlow = slow;slow = slow.next;}//System.out.println(preSlow.val);// 断开，分成两段preSlow.next = null;ListNode first = sortList(head);// 得到以排序好的前半段ListNode second = sortList(slow);// 得到以排序好的后半段ListNode dummy = new ListNode(-1);ListNode dummyCur = dummy;while(first!=null && second!=null){// 合并两半段if(first.val<second.val){dummyCur.next = first;first = first.next;}else if(second.val<=first.val){dummyCur.next = second;second = second.next;}dummyCur = dummyCur.next;}while(first != null){dummyCur.next = first;first = first.next;dummyCur = dummyCur.next;}while(second != null){dummyCur.next = second;second = second.next;dummyCur = dummyCur.next;}return dummy.next;    }}

Merge Sort:

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public static ListNode sortList(ListNode head) {if (head == null || head.next == null) {return head;}ListNode slow = head, fast = head, preSlow = head;while (fast != null && fast.next != null) {preSlow = slow;slow = slow.next;fast = fast.next.next;}preSlow.next = null;ListNode h1 = sortList(head);ListNode h2 = sortList(slow);return mergeList(h1, h2);}public static ListNode mergeList(ListNode h1, ListNode h2) {ListNode dummyHead = new ListNode(0);ListNode dummy = dummyHead;while (h1 != null && h2 != null) {if (h1.val <= h2.val) {dummy.next = h1;h1 = h1.next;} else {dummy.next = h2;h2 = h2.next;}dummy = dummy.next;}while (h1 != null) {dummy.next = h1;h1 = h1.next;dummy = dummy.next;}while (h2 != null) {dummy.next = h2;h2 = h2.next;dummy = dummy.next;}return dummyHead.next;}}

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