基于箱子排序对一堆n组卡片进行排序（C++单链表描述）

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#include<iostream>#include<algorithm> //STL中的算法#include<numeric>//标准库中的数学操作函数#include "chain.h"//对卡片进行排序//template<typename T>void cardSort(int n, chain<card>& L){//先对面值进行排序(有5种面值)for (int k = 1; k <= 3; ++k){int theBin;switch (k) {case 1:             //对面值排序theBin = 5;break;case 2:             //对样式排序theBin = 5;break;case 3:             //对组号排序theBin = n;break;}//先创建箱子chainNode<card>**bottom, **top;bottom = new chainNode<card>*[theBin];top = new chainNode<card>*[theBin];for (int b = 0; b < theBin; ++b)top[b] = NULL;//将链表的节点分配到箱子for (; L.firstNode != NULL; L.firstNode = L.firstNode->next){int i = 0;if (k == 1){float f = (L.firstNode->element).face; if(f==5)i = 0; else if(f==10)i = 1; else if (f ==20)i = 2; else if (f ==50)i = 3; else if (f ==100)i = 4;} if (k == 2){string s = (L.firstNode->element).deck;if (s == "a")i = 0;else if (s =="b")i = 1;else if (s=="c")i = 2;else if (s=="d")i = 3;else if (s=="e")i = 4;}if (k == 3)i = (L.firstNode->element).suit;if (top[i] == NULL)                //该面值只出现了一次top[i] = bottom[i] = L.firstNode;else                               //该面值出现了多次{bottom[i]->next = L.firstNode;bottom[i] = L.firstNode;}}//从箱子的节点中收集有序链表chainNode<card>*y = NULL;for (int i = 0; i < theBin; ++i)if (top[i] != NULL){if (y == NULL)             //搜集第一个非空箱子L.firstNode = top[i];else                       //搜集其他的箱子y->next = top[i];y = bottom[i];             //顺序连接}if (y != NULL)y->next = NULL;}for (chainNode<card>*p = L.firstNode; p != NULL; p = p->next)cout << (p->element).suit << " " << (p->element).deck << " " << (p->element).face << "  ";}int main(){//对卡片排序进行测试chain<card>money(15);//10张卡片//给这10张卡片赋值然后把他们插入链表，形成大小为10的链表string s[15] = { "e","b","c","a","b","d","c","e","a","b","d","c","e","a","b" };//卡片样式(假设卡片样式有a,b,c,d和e这5种)float f[15] = { 100,50,50,5,50,20,10,50,50,100,20,10,50,10,20 };        //卡片面值(假设面值有5，10,20,50和100这5种)int su[15] = { 9,8,1,6,6,5,3,2,2,1,5,3,2,2,1 };                     //卡片组号for (int i = 0; i<15; ++i){card cardRecord;cardRecord.suit = su[i];cardRecord.deck = s[i];cardRecord.face = f[i];money.insert(i, cardRecord);}cardSort(10,money);return 0;}
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