链表list容器中通过splice合并链表与merge的不同，及需要注意的问题

#include "stdafx.h"

#include <iostream>

#include <list>

#include <algorithm>

using namespace std;

int_tmain(int argc, _TCHAR* argv[])

{

list<int> c1,c2,c3,c4;

c1.push_back(3);

c1.push_back(6);

c2.push_back(2);

c2.push_back(4);

c3.push_back(5);

c3.push_back(1);

c4.push_back(40);

c4.push_back(41);

cout<<"c1="<<endl;

copy(c1.begin(),c1.end(),ostream_iterator<int>(cout,""));

cout<<endl;

cout<<"c2="<<endl;

copy(c2.begin(),c2.end(),ostream_iterator<int>(cout,""));

cout<<endl;

c2.splice(c2.begin(),c1);//将链表c1插入到链表c2的链头：第一种使用方式

cout<<"After splice c1 and c2>: c2=: "<<endl;

copy(c2.begin(),c2.end(),ostream_iterator<int>(cout,""));

cout<<endl;

cout<<"After splice c1 and c2: c1="<<endl;

copy(c1.begin(),c1.end(),ostream_iterator<int>(cout,""));

cout<<"可见splice合并后c1中没有内容了"<<endl;

cout<<"c3="<<endl;

copy(c3.begin(),c3.end(),ostream_iterator<int>(cout,""));

cout<<endl;

c2.splice(c2.begin(),c3,c3.begin());//将c3链表的头元素插入c2链表的头部：第二种使用方式

cout<<"After splice c2 and c3: c2=: "<<endl;

copy(c2.begin(),c2.end(),ostream_iterator<int>(cout,""));

cout<<endl;

cout<<"After splice c3="<<endl;

copy(c3.begin(),c3.end(),ostream_iterator<int>(cout,""));

cout<<endl;

cout<<"可见splice后，c3的头元素不在了"<<endl;

cout<<"c4="<<endl;

copy(c4.begin(),c4.end(),ostream_iterator<int>(cout,""));

cout<<endl;

c2.splice(c2.begin(),c4,c4.begin(),c4.end());//将链表c4从开始到结束都合并到c2开始的位置：第三种使用方式

cout<<"After splice c2 and c4: c2=: "<<endl;

copy(c2.begin(),c2.end(),ostream_iterator<int>(cout,""));

cout<<endl;

cout<<"After splice c4="<<endl;

copy(c4.begin(),c4.end(),ostream_iterator<int>(cout,""));

cout<<endl;

cout<<"可见splice合并后c4中没有内容了"<<endl;

return 0;

}

执行结果：



可见：splice与merge最大的不同时，不用排序，也不要求原始链表有序。相同点是，被合并的链表或元素将消失。