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* t = malloc(sizeof(struct ListNode));
    t->val = -1001;
    t->next = pHead1;
    struct ListNode* l1 = t;
    struct ListNode* l2 = pHead2;
    while(l1->next != NULL && l2 != NULL) {
        if(l2->val >= l1->val && l2->val <= l1->next->val) {
            printf("l2->val is %d and l1->next->val is %d\n", l2->val, l1->next->val);
            struct ListNode* l1n = l1->next;
            printf("l1->next val is %d\n", l1n->val);
            l1->next = l2;
            while(l2->next != NULL && l2->next->val <= l1n->val) {
                l2 = l2->next;
            }
            l1 = l2;
            l2 = l2->next;
            l1->next = l1n;
            l1 = l1n;
        } else {
            l1 = l1->next;
        }
    }
    if(l2 != NULL) {
        l1->next = l2;
    }
    return t->next;
}