[刷题]Sort List

[LintCode]Sort List

/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     } * } */ public class Solution {    /**     * @param head: The head of linked list.     * @return: You should return the head of the sorted linked list,                    using constant space complexity.     */    public ListNode sortList(ListNode head) {          // 2015-5-24 O(nlogn) 分治法 递归 类似归并排序        // exit condition        if (head == null || head.next == null) {            return head;        }                ListNode mid = findMid(head);        ListNode listB = sortList(mid.next);        mid.next = null;        ListNode listA = sortList(head);        // listB 不长于 listA                return mergeList(listA, listB);            }        private ListNode findMid(ListNode head) {        if (head == null) {            return head;        }        ListNode slow = head;        ListNode fast = head;        while (fast.next != null && fast.next.next != null) {            slow = slow.next;            fast = fast.next.next;        }        return slow;    }        private ListNode mergeList(ListNode listA, ListNode listB) {        ListNode dummy = new ListNode(0);        ListNode tail = dummy;        while (listA != null && listB != null) {            if (listA.val < listB.val) {                tail.next = listA;                listA = listA.next;            } else {                tail.next = listB;                listB = listB.next;            }            tail = tail.next;        }        if (listA != null) {            tail.next = listA;        }        if (listB != null) {            tail.next = listB;        }        return dummy.next;    }}