"简简单单"
2025-10-09 21:12:37
发布于:四川
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终于学会链表了!!!
#include<bits/stdc++.h>
using namespace std;
struct A{
int d;
int next;
};
int main(){
A a[105];
int n;
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i].d;
a[i].next=i+1;
}
int k,ki;
cin>>k>>ki;
a[ki].next=n+1;
a[n+1].next=ki+1;
a[n+1].d=k;
int i=1,ai=1;
while(i<=n+1){
cout<<a[ai].d<<" ";
ai=a[ai].next;
i++;
}
return 0;
}
全部评论 2
猪妞行为
2025-11-08 来自 四川
1你这是个迪奥的链表,链表解法是这样的
#include <iostream> using namespace std; struct node{ int val; node* next; }; void show_list(node* head){ //遍历输出链表中所有元素 node* cur = head; while(cur != nullptr){ cout<<cur->val<<" "; cur = cur->next; } } int main(){ int n; cin>>n; node* dummy = new node(); dummy->next = nullptr; node* tail = dummy; for(int i=0;i<n;i++){ int num; cin>>num; node* nd = new node(); nd->val = num; nd->next = nullptr; tail->next = nd; tail = nd; }int m,q; cin>>m>>q; node* cur = dummy->next; while(cur != nullptr){ if(cur->val == m){ node* newp = new node(); newp->val = q; newp->next = cur->next; cur->next = newp; cur = cur->next; } cur = cur->next; } show_list(dummy->next); return 0; }2026-07-30 来自 山东
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