[LintCode] Merge Two Sorted Lists [附Recursion solution]

Merge Two Sorted Lists

Merge two sorted (ascending) linked lists and return it as a new sorted list. The new sorted list should be made by splicing together the nodes of the two lists and sorted in ascending order.

Example

Given 1->3->8->11->15->null, 2->null , return 1->2->3->8->11->15->null.

 

SOLUTION1:

链表在merge的时候其实要有优势的，它不需要额外空间，不想array，要重新开一个空间放array。操作的时候也很简单，弄一个dummy，然后把小的直接连上就行了。此题属于基本操作，在其他题里完全可以作为function调用。

代码：

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param ListNode l1 is the head of the linked list
     * @param ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null){
            return l2;
        }
        if (l2 == null){
            return l1;
        }
        ListNode dummy = new ListNode(0);
        ListNode node = dummy;
        while (l1 != null && l2 != null){
            if (l1.val < l2.val){
                node.next = new ListNode(l1.val);
                l1 = l1.next;
                node = node.next;
            } else {
                node.next = new ListNode(l2.val);
                l2 = l2.next;
                node = node.next;
            }
        }
        while (l1 != null){
            node.next = new ListNode(l1.val);
            l1 = l1.next;
            node = node.next;
        }
        while (l2 != null){
            node.next = new ListNode(l2.val);
            l2 = l2.next;
            node = node.next;
        }
        return dummy.next;
    }
}

 

SOLUTION2:

当然链表也可以使用recursion解决，在recursion的过程中由于callbacl stack的使用，导致空间复杂度退化为O(n) n为两个链表最长的长度。但是在写法上实在是太简化了，所以也要学会使用。

Recursion:

1, Define subproblem;

2, Find recursion tree;

3, Define base case;

本题中：

1, Define subproblem;  l1.next / l2.next (P.S, 链表问题都是砍“头”，即子问题变为cur.next的问题）

2, Find recursion rule;  l1.next / l2.next 将砍掉的两个头的next指针接到新返回的newHead上

3, Define base case;  一个为null，返回另一个

public class Solution {
    /*
     * @param l1: ListNode l1 is the head of the linked list
     * @param l2: ListNode l2 is the head of the linked list
     * @return: ListNode head of linked list
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        //base case
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }
        
        if (l1.val <= l2.val) { 
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}