148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.

这道题要求用nlogn的时间复杂度去做，代码如下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null||head.next==null) return head;//corner case:if the List is null or the length of List is just one.
        ListNode fast = head,slow = head,prev = null;
        while(fast!=null&&fast.next!=null){//find the middle of the listnode
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        //divide the List into two parts
        prev.next = null;
        //recursive the sortList
        ListNode l1=sortList(head);
        ListNode l2=sortList(slow);
        //conquer two parts together
        return combine(l1,l2);
    }
    public ListNode combine(ListNode l1,ListNode l2){
        ListNode l = new ListNode(0),p = l;
        while(l1!=null&&l2!=null){
            if(l1.val<=l2.val){
                p.next = l1;
                l1 = l1.next;
            }else{
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
            if(l1!=null){
                p.next = l1;
            }
            if(l2!=null){
                p.next = l2;
            }
        }
        return l.next;
    }
}
// 

本题是典型的merge sort，只不过输入的数据是链表而不是数组了，那么我们需要每次找到中间的位置，要用到快慢指针，所以每次寻找中间位置的时间复杂度是O（n），总共需要运行logn次，所以总的时间复杂度是nlongn