题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pHead1 ListNode类 
     * @param pHead2 ListNode类 
     * @return ListNode类
     */
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        // write code here
        ListNode *s1;
        ListNode *s2;
        ListNode *tmps1;
        ListNode *tmps2;
        ListNode *s;

        if(pHead1 == nullptr || pHead2 == nullptr)
            return pHead1 ? pHead1 : pHead2;

        if(pHead1->val <= pHead2->val){
            s1 = pHead1;
            s2 = pHead2;
            s = pHead1;
        }else{
            s1 = pHead2;
            s2 = pHead1;
            s = pHead2;
        }
	  
        tmps1 = s1;
        tmps2 = s2;
	  
        while(s2 != nullptr){
            while(s1 != nullptr && s1->val <= s2->val){
                    tmps1 = s1;         
                    s1 = s1->next;
                }
                       
            tmps1->next = s2; 
            tmps1 = tmps1->next;                    
            tmps2 = s2;
 
            s2 = s2->next;
            tmps2->next = s1;                     
        }

        return s;
    }
};