lintcode：合并两个排序链表

题目：

合并两个排序链表

将两个排序链表合并为一个新的排序链表

 样例

给出 1->3->8->11->15->null，2->null，

返回 1->2->3->8->11->15->null。

解题：

数据结构中的书上说过，可解，异步的方式移动两个链表的指针，时间复杂度O(n+m)

Java程序：

/**
 * 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) {
        // write your code here
        if(l1==null && l2!=null)
            return l2;
        if(l1!=null && l2==null)
            return l1;
        if(l1==null && l2==null)
            return null;
        ListNode head = new ListNode(0);
        ListNode current = head;
        
      
        while(l1!=null && l2!=null){
                if(l1.val<=l2.val){
                    current.next = l1;
                    current = current.next;
                    l1 = l1.next;
                }else{
                    current.next = l2;
                    current = current.next;
                    l2 = l2.next;
                }
            }

        if(l1!=null)
            current.next= l1;
        if(l2!=null)
            current.next=l2;
        return head.next;
    }
}

总耗时: 13348 ms

Python程序：

"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param two ListNodes
    @return a ListNode
    """
    def mergeTwoLists(self, l1, l2):
        # write your code here
        if l1==None:
            return l2
        if l2==None:
            return l1
        if l1==None and l2==None:
            return None
        head = ListNode(0)
        p = head
        while l1!=None and l2!=None:
            if l1!=None and l2!=None:
                if l1.val<= l2.val:
                    p.next = l1
                    p = p.next
                    l1 = l1.next
                else:
                    p.next = l2
                    p = p.next
                    l2 = l2.next
            if l1==None:
                p.next = l2
                break
            if l2==None:
                p.next = l1
                break
        return head.next

总耗时: 2032 ms

 参考剑指OfferP117

利用递归的思想

小的节点链接，下一次递归

递归的好处是不用单独搞个节点当作头节点了，通俗点说是许多头节点连接起来的，最终我们返回的是第一个头节点

/**
 * 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) {
        // write your code here
        if(l1==null && l2!=null)
            return l2;
        if(l1!=null && l2==null)
            return l1;
        if(l1==null && l2==null)
            return null;
        ListNode MergeHead = null;
        if(l1.val < l2.val){
            MergeHead = l1;
            MergeHead.next = mergeTwoLists(l1.next,l2);
        }else{
            MergeHead = l2;
            MergeHead.next = mergeTwoLists(l1,l2.next);
        }
        return MergeHead;
        
    }
}

总耗时: 14047 ms

"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param two ListNodes
    @return a ListNode
    """
    def mergeTwoLists(self, l1, l2):
        # write your code here
        if l1==None:
            return l2
        if l2==None:
            return l1
        if l1==None and l2==None:
            return None
        head = None
        if l1.val< l2.val:
            head = l1
            head.next = self.mergeTwoLists(l1.next,l2)
        else:
            head = l2
            head.next = self.mergeTwoLists(l1,l2.next)
        return head 

总耗时: 2403 ms