[LeetCode 148]Sort List

题目链接：sort-list

/** * Sort a linked list in O(n log n) time using constant space complexity. * */public class SortList {public class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}//解法一 ： 归并法//15 / 15 test cases passed.//Status: Accepted//Runtime: 310 ms//Submitted: 0 minutes ago//时间复杂度O(n * log(n)) 空间复杂度 O(1)    public ListNode sortList(ListNode head) {    if(head == null || head.next == null) {    return head;    }    ListNode slow = head;    ListNode fast = head;    //找到链表的中点位置    while(fast.next != null && fast.next.next != null) {    fast = fast.next.next;    slow = slow.next;    }    fast = slow.next;    //断开链表    slow.next = null;        ListNode list1 = sortList(head);    ListNode list2 = sortList(fast);    return mergeList(list1, list2);    }    //合并两链表    public ListNode mergeList(ListNode head1, ListNode head2) {    ListNode head = new ListNode(-1);    ListNode last = head;    while(head1 != null && head2 != null) {    if(head1.val <= head2.val) {    last.next = head1;    head1 = head1.next;    } else {    last.next = head1;    head1 = head1.next;}    last = last.next;    }    if(head1 != null)    last.next = head1;    if(head2 != null)    last.next = head2;    return head.next;    }//解法二//15 / 15 test cases passed.//Status: Accepted//Runtime: 709 ms //Submitted: 0 minutes ago// 对排序好的链表设置一个中点指针，如果待排的节点大于中点指针的值，则从中点开始寻找插入点，否则从表头开始查找    //其实也没降低时间复杂度 和暴力插排一样 仍为O(n*n) 空间复杂度 O(1)    public ListNode sortList1(ListNode head) {        ListNode preHead = new ListNode(-1);        ListNode last = null;        ListNode cur = null;                int preNum = 0, postNum = 0;                while(head != null) {        ListNode headNext = head.next;        if(last != null && last.val <= head.val) {        postNum ++;        cur = last;        } else {        preNum ++;        cur = preHead;}                while(cur.next != null) {if(cur.next.val >= head.val) {break;}cur = cur.next;}if(last == null) {last = head;}    ListNode curNext = cur.next;cur.next = head;cur.next.next = curNext;        head = headNext;        if(postNum > preNum) {last = last.next;}        }    return preHead.next;    }public static void main(String[] args) {// TODO Auto-generated method stub}}