算法和数据结构就是编程的一个重要部分，你若失掉了算法和数据结构，你就把一切都失掉了。
开始理解错题目，但是创建双向链表的思路可行。
#include<iostream> #include <stdlib.h> using namespace std; //双向循环链表 typedef int datatype; //方便修改 //当然也可以写成模板来适应更多的数据类型 struct dclist{ datatype data;//数据定义 struct dclist *prior; struct dclist *next;//前驱和后继指针 }; class Joseph { public: int getResult(int n, int m) { dclist *L = (struct dclist*)malloc(sizeof(struct dclist)); L->next = L->prior = NULL; dclist *r = L; dclist *p; for (int i = 1; i <= n; i++){ p = (struct dclist*)malloc(sizeof(struct dclist)); p->data = i; p->prior = r; p->next = r->next; r->next = p; r = p; } r->next = L->next; L->next->prior = r; if (m>0) do{ L = L->next; } while (--m); int t = L->data; return t; } }; int main(){ int n, m; cin >> n >> m; Joseph tlist; int t = tlist.getResult(n, m); cout << t; }
后面的是正确答案
#include<iostream> #include <stdlib.h> using namespace std; //双向循环链表 typedef int datatype; //方便修改 //当然也可以写成模板来适应更多的数据类型 struct dclist{ datatype data;//数据定义 struct dclist *prior; struct dclist *next;//前驱和后继指针 }; class Joseph { public: int getResult(int n, int m) { dclist *L = (struct dclist*)malloc(sizeof(struct dclist)); L->next = L->prior = NULL; dclist *r = L; dclist *p; for (int i = 1; i <= n; i++){ p = (struct dclist*)malloc(sizeof(struct dclist)); p->data = i; p->prior = r; p->next = r->next; r->next = p; r = p; } r->next = L->next; L->next->prior = r; //删除操作 while (L->next != L){ for (int k = 1; k < m; k++) //因为要保留L，删除p结点，所以这里循环m-1次 L = L->next; p = L->next; L->next = p->next; free(p);//删除p } // int t = L->data; return t; } }; int main(){ int n, m; cin >> n >> m; Joseph tlist; int t = tlist.getResult(n, m); cout << t; }