题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 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
 */
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
    // write code here
    struct ListNode* doge=(struct ListNode*)malloc(sizeof(struct ListNode));
    doge->next=pHead1;
    struct ListNode*prev=doge;
    while(pHead1&&pHead2){
        if(pHead1->val>=pHead2->val){
            prev->next=pHead2;
            pHead2=pHead2->next;
        }
        else{
            prev->next=pHead1;
            pHead1=pHead1->next;
        }
        prev=prev->next;
    }
if(pHead1!=NULL){
    prev->next=pHead1;
}
if(pHead2!=NULL){
    prev->next=pHead2;
}
 return doge->next;
}