题解 | #合并两个排序的链表#

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        ListNode* ans = new ListNode(0), *cur = ans;
        while (pHead1 && pHead2) {  // 两个链表均不为空，比较大小
            if (pHead1->val <= pHead2->val) {
                cur->next = pHead1;
                pHead1 = pHead1->next;
            }
            else {
                cur->next = pHead2;
                pHead2 = pHead2->next;
            }
            cur = cur->next;
        }
        if (pHead1) { // 第一个链表剩余部分不为空
            cur->next = pHead1;
        }
        if (pHead2) { // 第二个链表剩余部分不为空
            cur->next = pHead2;
        }
        return ans->next;
    }
};