[Leetcode][JAVA] Merge Two Sorted Lists & Sort List

Merge Two Sorted Lists:

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

两个排好序的链表拼接，只要用两个指针遍历链表即可. 可以借助一个helper指针来进行开头的合并。如果有一个遍历完，则直接把另一个链表指针之后的部分全部接在后面：

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode helper = new ListNode(0);
        ListNode cur1 = l1;
        ListNode cur2 = l2;
        ListNode cur = helper;
        while(cur1!=null || cur2!=null)
        {
            if(cur1==null)
            {
                cur.next = cur2;
                break;
            }
            if(cur2==null)
            {
                cur.next = cur1;
                break;
            }
            if(cur1.val<=cur2.val)
            {
                cur.next = cur1;
                cur1=cur1.next;
            }
            else
            {
                cur.next = cur2;
                cur2=cur2.next;
            }
            cur=cur.next;
        }
        return helper.next;
    }

 

Sort List:

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

有了Merged Two Sorted Lists的基础，这题就很好写了。提到O(nlogn)的排序算法，马上想到快排和合并排序，不过快排不稳定而且对于链表来说交换不方便，所以还是决定采用merge sort.

对于链表，merge sort中的split过程可以使用快慢指针实现，比较简单易懂：

 

public ListNode sortList(ListNode head) {
        return mergeSort(head);
    }
    
    public ListNode mergeSort(ListNode start)
    {
        if(start==null) return null;
        if(start.next==null) return start;
        
        ListNode slow = start;
        ListNode fast = start;
        while(fast!=null && fast.next!=null && fast.next.next!=null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode second = slow.next;
        slow.next = null;
        return mergeTwoLists(mergeSort(start), mergeSort(second));
    }
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode helper = new ListNode(0);
        ListNode cur1 = l1;
        ListNode cur2 = l2;
        ListNode cur = helper;
        while(cur1!=null || cur2!=null)
        {
            if(cur1==null)
            {
                cur.next = cur2;
                break;
            }
            if(cur2==null)
            {
                cur.next = cur1;
                break;
            }
            if(cur1.val<=cur2.val)
            {
                cur.next = cur1;
                cur1=cur1.next;
            }
            else
            {
                cur.next = cur2;
                cur2=cur2.next;
            }
            cur=cur.next;
        }
        return helper.next;
    }