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。