[LeetCode]80. Merge Two Sorted Lists合并两个排序链表

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.

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解法1：递归。首先比较头节点大小，若l2->val>l1->val，则返回mergeTwoLists(l2,l1)；否则（1）如果l1->next!=NULL，则比较l1->next->val与l2->val的大小，这样可以确定前两个元素的顺序，

/**
 * 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) {
            if (l1->next != NULL && l1->next->val <= l2->val){
                ListNode* node = mergeTwoLists(l1->next->next, l2);
                l1->next->next = node;
            }
            else if (l1->next != NULL && l1->next->val > l2->val) {
                ListNode* node = mergeTwoLists(l1->next, l2->next);
                l2->next = node;
                l1->next = l2;
            }
            else
                l1->next = l2;
            return l1;
        }
        else
            return mergeTwoLists(l2, l1);
    }
};

解法2：迭代。新建一个指针help，然后遍历l1和l2，将二者中较小的链接到help后面。最后返回help的下一个节点作为头节点。注意两个链表可能长度不一样，要将长链表最后剩下的元素链接到新链表中。

/**
 * 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;
        ListNode* help = new ListNode(0);
        ListNode* head = help;
        while (l1 != NULL && l2 != NULL) {
            if (l1->val <= l2->val) {
                help->next = l1;
                l1 = l1->next;
            }
            else {
                help->next = l2;
                l2 = l2->next;
            }
            help = help->next;
        }
        if (l1 != NULL) help->next = l1;
        if (l2 != NULL) help->next = l2;
        return head->next;
    }
};