21. Merge Two Sorted Lists【easy】

21. Merge Two Sorted Lists【easy】

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.

解法一：

/**
 * 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) {
        if (l1 == NULL || l2 == NULL) {
            return l1 ? l1 : l2;
        }
        
        ListNode * dummy = new ListNode(INT_MIN);
        ListNode * temp = dummy;
        
        while (l1 && l2) {
            if (l1->val > l2->val) {
                dummy->next = l2;
                l2 = l2->next;
            }
            else {
                dummy->next = l1;
                l1 = l1->next;
            }
            
            dummy = dummy->next;
        }
        
        if (l1 || l2) {
            dummy->next = l1 ? l1 : l2;
        }
        
        return temp->next;
    }
};

由于最后是弄到list1中，但是我们不知道list1还是list2的第一个元素关系，最后结果的list1中的头结点可能会改变，所以需要引入dummy节点。

解法二：

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;
        }
}

参考了@yangliguang 的代码

解法三：

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        
        ListNode handler;
        if(l1.val < l2.val) {
            handler = l1;
            handler.next = mergeTwoLists(l1.next, l2);
        } else {
            handler = l2;
            handler.next = mergeTwoLists(l1, l2.next);
        }
        
        return handler;
    }
}

参考了@RunRunCode 的代码

解法二和解法三都是递归，还没有完全弄明白……