21 Merge Two Sorted Arrays

两种思路 一种是recursion，每次处理两个数组的头，然后循环调用自己，处理后面的部分。空间复杂度 O(n+m) 时间复杂度 O(n+m) 第二种是Iteration, 时间复杂度O(n+m), 空间复杂度O(1)

Python

```python

Definition for singly-linked list.

class ListNode(object):

def init(self, x):

self.val = x

self.next = None

class Solution(object): def mergeTwoLists(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ if l1 is None: return l2 elif l2 is None: return l1 elif l1.val < l2.val: l1.next = self.mergeTwoLists(l1.next, l2) return l1 else: l2.next = self.mergeTwoLists(l1,l2.next) return l2

```python
class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        prehead = ListNode(-1)

        prev = prehead
        while l1 and l2:
            if l1.val <= l2.val:
                prev.next = l1
                l1 = l1.next
            else:
                prev.next = l2
                l2 = l2.next
            prev = prev.next
        prev.next = l1 if l1 else l2
        return prehead.next

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        if(l1.val < l2.val){
            l1.next = this.mergeTwoLists(l1.next, l2);
            return l1;
        }
        else{
            l2.next = this.mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode p = dummy;
        while (l1 != null||l2 != null)
        {
            if (l1 == null){
                p.next = l2;
                return dummy.next;
            }
            if (l2 == null){
                p.next = l1;
                return dummy.next;
            }
            if(l1.val < l2.val){
                p.next = new ListNode(l1.val);
                l1 = l1.next;
            }
            else{
                p.next = new ListNode(l2.val);
                l2 = l2.next;
            }
            p = p.next;
        }
        return dummy.next;
    }
}