LeetCode 148. Sort List以及排序

排序算法： 时间复杂度为 O(nlgn）:
quick sort (空间复杂度O(1)), merge sort(空间复杂度O(n)), heap sort (空间复杂度O(1)) 时间复杂度为 O(n): bucket sort, radix sort, 基于比较的排序，时间复杂度一般为O(nlgn）

Question

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

Explanation

这道题主要要掌握几种sort的方法怎么写，还有相关的复杂度分析。

Code

1. Merge Sort: 时间复杂度为O(nlogn），空间复杂度O(n)

   public class Solution {
       public ListNode sortList(ListNode head) {
           if (head == null || head.next == null) return head;
           // Merge Sort
           ListNode mid = findMiddle(head);
           ListNode right = sortList(mid.next);
           mid.next = null;
           ListNode left = sortList(head);
           return mergeTwoLists(left, right);
   
   
       }
   
       public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
           ListNode a = l1;
           ListNode b = l2;
           if (l1 == null && l2 == null) return null;
           ListNode curr = new ListNode(0);
           ListNode dummy= curr;
           while(a != null && b != null) {
               if (a.val < b.val) {
                   dummy.next = new ListNode(a.val);
                   a = a.next;
               } else {
                   dummy.next = new ListNode(b.val);
                   b = b.next;
               }
               dummy = dummy.next;
           }
           if(a!= null) dummy.next = a;
           else dummy.next = b;
           return curr.next;
       }
   
   
   
       private ListNode findMiddle(ListNode head) {
           ListNode walker = head;
           ListNode runner = head.next;
           while(runner!= null && runner.next!=null) {
               runner = runner.next.next;
               walker = walker.next;
           }
           return walker;
       }
   }

2. Quick Sort

public class Solution {
    public ListNode sortList(ListNode head) {
        //Quick Sort
        if(head == null || head.next == null) return head;
        ListNode mid = findMedian(head);//O(n)
        ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
        ListNode rightDummy = new ListNode(0), rightTail = rightDummy;
        ListNode middleDummy = new ListNode(0), middleTail = middleDummy;
        while (head != null) {
            if (head.val < mid.val) {
                leftTail.next = head;
                leftTail = head;
            } else if (head.val > mid.val) {
                rightTail.next = head;
                rightTail = head;
            } else {
                middleTail.next = head;
                middleTail = head;
            }
            head = head.next;
        }
        leftTail.next = null;
        middleTail.next = null;
        rightTail.next = null;
        ListNode left = sortList(leftDummy.next);
        ListNode right = sortList(rightDummy.next);
        return concat(left, middleDummy.next, right);

    }
    private ListNode findMedian(ListNode head) {
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    private ListNode concat(ListNode left, ListNode middle, ListNode right) {
        ListNode dummy = new ListNode(0), tail = dummy;

        tail.next = left; tail = getTail(tail);
        tail.next = middle; tail = getTail(tail);
        tail.next = right; tail = getTail(tail);
        return dummy.next;
    }
    private ListNode getTail(ListNode head) {
        if (head == null) {
           return null;
        }

        while (head.next != null) {
            head = head.next;
        }
        return head;
    }


}