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。