合并两个单调不递减链表

Question

这么写为什么不对啊 有希望改嘛 错在哪里了 初学者自学的 别提醒我30个字了

Answer 1

#include <bits/stdc++.h>

using namespace std;

typedef int ElemType;

typedef struct Link {
  ElemType Elem;
  struct Link *next;
} link;

link *CreatLink(int n) {
  link *head = (link *)malloc(sizeof(link));
  head->next = NULL;
  //创立头节点
  link *temp = head;
  //创立头指针指向头节点
  while (n--) {
    link *a = (link *)malloc(sizeof(link));
    scanf("%d", &a->Elem);
    a->next = NULL;
    temp->next = a;
    temp = a;
  }
  return head;
}

link *merge(link *l1, link *l2) {
  if (l1 == NULL)
    return l2;
  if (l2 == NULL)
    return l1;
  l1 = l1->next;
  l2 = l2->next;
  link *head = (link *)malloc(sizeof(link));
  head->next = NULL;
  //新头节点
  link *tail = head;
  //新头指针
  while (l1 && l2) {
    if (l1->Elem <= l2->Elem) {
      tail->next = l1;
      l1 = l1->next;
    } else {
      tail->next = l2;
      l2 = l2->next;
    }
    tail = tail->next;
  }
  //循环操作

  if (l1)
    tail->next = l1;
  if (l2)
    tail->next = l2;
  //最后必有一个原链表还剩下一串链表

  return head;
}

int main() {
  link *l1, *l2, *l3;
  int n1, n2;

  cout << "输入A链表表长及元素\n";
  scanf("%d", &n1);
  l1 = CreatLink(n1);

  cout << "输入B链表表长及元素\n";
  scanf("%d", &n2);
  l2 = CreatLink(n2);

  l3 = merge(l1, l2);
  cout << "合并后的C表为:\n";

  link *p = l3->next;
  while (p) {
    printf("%d ", p->Elem);
    p = p->next;
  }

  return 0;
}