题解 | 牛牛的单链表求和
牛牛的单链表求和
https://www.nowcoder.com/practice/a674715b6b3845aca0d5009bc27380b5
#include <iostream> #include <string> using namespace std; struct list{ int num; list* next; }; int main() { int n; cin>>n; //创建链表 list* head = (list*)malloc(sizeof(list)); list* pCurrent=head; head->next=nullptr; for (int i=0;i<n;i++) { list* tmp=(list*)malloc(sizeof(list)); cin>>tmp->num; tmp->next=nullptr; pCurrent->next=tmp;//连接 pCurrent=pCurrent->next; } //重置pCurrent pCurrent=head->next; int sum=0; while (pCurrent!=nullptr) { sum+=pCurrent->num; pCurrent=pCurrent->next; } cout<<sum; } // 64 位输出请用 printf("%lld")