21. Merge Two Sorted Lists

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本文介绍了一种将两个已排序的链表合并为一个新排序链表的方法。提供了递归和非递归两种实现方式，并详细展示了每种方法的具体步骤。通过比较不同实现方式的空间复杂度，帮助读者理解各种解决方案的特点。

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.

// recursive solution, time O(m+n), much more space complexity

/**
 * 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)
        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; 
        }
        
     
    }
};

my 非递归法

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1 == NULL)
        return l2;
        if(l2 == NULL)
        return l1;
        ListNode* p = l1;
        ListNode* q = l2;
        ListNode* x = (l1->val <= l2->val) ? l1 :l2;
        ListNode* c;
        
        
        while(p&&q)
        {
            if(p->val <=q->val)
            {
                c->next = p;
                c = c->next;
                p = p->next;
            }
            else
            {
                c->next = q;
                c = c->next;
                q = q->next;
            }
            
        }
        if(!p)
        {
            c->next = q;
            
        }
        if(!q)
        {
            c->next = p;
            
        }
        return x;
        
        
     
    }
};

// merge sort, adding while comparing, time O(m+n)
class Solution_MergeSort {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode dummy(0);                                     //头！
        ListNode *curr = &dummy;
        while (l1 && l2)
        {
            if (l1->val < l2->val)
            {
                curr->next = l1;
                curr = l1;
                l1 = l1->next;
            }
            else
            {
                curr->next = l2;
                curr = l2;
                l2 = l2->next;
            }
        }

        if (!l1)
        {
            curr->next = l2;
        }

        if (!l2)
        {
            curr->next = l1;
        }

        return dummy.next;
    }
};