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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

---

归并两个有序链表，要求使用原有链表中的节点，所以只需要依次比较大小，改变原有节点的指向。 代码如下

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
    {
        ListNode *head = new ListNode(0), *l;
        l = head;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                l->next = l1;
                l1 = l1->next;
            }
            else {
                l->next = l2;
                l2 = l2->next;
            }
            l = l->next;
        }
        if (l1)
            l->next = l1;
        else if (l2)
            l->next = l2;
        return head->next;
    }
};

本代码提交AC，用时8MS。

二刷。循环的时候用或运算也可以，代码更简洁。

class Solution {
public:
	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
		if (l1 == NULL)return l2;
		ListNode* ans = new ListNode(0);
		ListNode* root = ans;
		while (l1 || l2) {
			int v1 = l1 ? l1->val : INT_MAX;
			int v2 = l2 ? l2->val : INT_MAX;
			if (v1 != INT_MAX && v1 <= v2) {
				root->next = l1;
				root = root->next;
				l1 = l1->next;
			}
			else if (v2 != INT_MAX && v2 < v1) {
				root->next = l2;
				root = root->next;
				l2 = l2->next;
			}
		}
		return ans->next;
	}
};

本代码提交AC，用时8MS。