合并两个有序递增链表
题目如下：
代码如下：
1 #include <stdio.h> 2 #include <stdlib.h> 3 4 typedef struct ListNode { 5 int val; 6 struct ListNode *next; 7 } ListNode_t; 8 9 struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) 10 { 11 struct ListNode* vhead;//新建一个头结点 12 struct ListNode *p = vhead;//用一个指针指向该头结点 13 while (pHead1 && pHead2)//两个链表都未比较完 14 { 15 if (pHead1->val <= pHead2->val) //表1的值更小 16 { 17 p->next = pHead1; //先将结点连接到新表，再让原表指针后移一位，二者顺序不可换 18 pHead1 = pHead1->next; 19 } 20 else 21 { 22 p->next = pHead2; 23 pHead2 = pHead2->next; 24 } 25 p = p->next; //加入一个新结点后，新表指针也后移一位 26 } 27 p->next = pHead1 ? pHead1 : pHead2; //都比较完后，哪个表非空则直接加入到新表中 28 return vhead->next; //返回第一个结点 29 } 30 31 int main() 32 { 33 // 链表1：1->20->95 34 ListNode_t * first_1; 35 ListNode_t * second_1; 36 ListNode_t * third_1; 37 first_1 = (ListNode_t*)malloc(sizeof(ListNode_t)); 38 second_1 = (ListNode_t*)malloc(sizeof(ListNode_t)); 39 third_1 = (ListNode_t*)malloc(sizeof(ListNode_t)); 40 first_1->val = 1; 41 second_1->val = 20; 42 third_1->val = 95; 43 first_1->next = second_1; 44 second_1->next = third_1; 45 third_1->next = NULL; 46 47 // 链表2：2->4->6 48 ListNode_t * first_2; 49 ListNode_t * second_2; 50 ListNode_t * third_2; 51 first_2 = (ListNode_t*)malloc(sizeof(ListNode_t)); 52 second_2 = (ListNode_t*)malloc(sizeof(ListNode_t)); 53 third_2 = (ListNode_t*)malloc(sizeof(ListNode_t)); 54 first_2->val = 2; 55 second_2->val = 4; 56 third_2->val = 6; 57 first_2->next = second_2; 58 second_2->next = third_2; 59 third_2->next = NULL; 60 61 ListNode_t* curr = Merge(first_1,first_2); 62 63 while(curr!=NULL) 64 { 65 printf("curr value = %d\n",curr->val); 66 curr = curr->next; 67 } 68 69 return 0; 70 }
传入链表1： 1->20->95
传入链表2： 2->4->6
期望实验结果：1->2->4->6->20->95
验证结果如下：
