【LeetCode】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.

题意：合并两个已排序的链表

题不难，但是有很多需要注意的细节，直接贴代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *tmp,*p1,*p2;
    if(l1==NULL){
        tmp=l1;
        l1=l2;
        l2=tmp;
    }
    if(l1==NULL)
        return l1;
    p1=l1;
    p2=l2;
    while(l1!=NULL){
        if(l2!=NULL&&l1->val<=l2->val){
            while(l1->next!=NULL&&l1->next->val<=l2->val) 
                l1=l1->next;
            tmp=l2;
            l2=l2->next;
            tmp->next=l1->next;
            l1->next=tmp;
            l1=l1->next;
        }
        else if(l2!=NULL&&l1->val>l2->val){
            tmp=l1;
            l1=l2;
            l2=tmp;
            p1=l1;
        }
        else if(l2==NULL)
            return p1;
    }
    return p1;
}

 感觉写的好乱，重新写一遍：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *tmp,*p1;
    if(l1==NULL&&l2==NULL)
        return NULL;
    if(l1==NULL)
        return l2;
    if(l2==NULL)
        return l1;
    if(l1->val>l2->val){
        tmp=l1;
        l1=l2;
        l2=tmp;
    }
    p1=l1;
    while(l1!=NULL){
        if(l2!=NULL&&l1->val<=l2->val){
            while(l1->next!=NULL&&l1->next->val<=l2->val) 
                l1=l1->next;
            tmp=l2;
            l2=l2->next;
            tmp->next=l1->next;
            l1->next=tmp;
        }
        else if(l2==NULL)
            return p1;
    }
    return p1;
}