21. 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.

思路1：这是一个Merge操作的实现，先搞都存在的部分。再搞各自多出来的部分。

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode newHead = new ListNode(0);
        ListNode prev = newHead;
        while (l1!=null && l2!=null){
            if (l1.val<l2.val){
                prev.next = l1;
                l1 = l1.next;
            }else{
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }

        if (l1==null){
            prev.next = l2;
        }        
        if (l2==null){
            prev.next = l1;
        }

        return newHead.next;
    }
}

思路2：递归。更高级和简明

public 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 = mergeTwoLists(l1.next, l2);
            return l1;
        } else{
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

C++ Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* newHead = new ListNode(0);
        ListNode* prev = newHead;
        while (l1 && l2){
            if (l1->val < l2->val){
                prev->next = l1;
                l1 = l1->next;
                prev = prev->next;
            }else{
                prev->next = l2;
                l2 = l2->next;
                prev = prev->next;
            }
        }
        if (l1){
            prev->next = l1;
        }
        if (l2){
            prev->next = l2;
        }

        return newHead->next;
    }
};