nullnull计算两个字符串链表中的共同数据项，需要考虑重复选项的情况

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    Problem

    每日一道理
谁说人与人隔着遥远的重洋，谁说心与心设着坚固的堤防？十六岁的鸟儿飞上天空，总会找到落脚的枝头。

/*

// Given two lists of strings build a new list that has all strings that appear in both the original lists. If the same string appears more than once output it as many times as it appears in both lists 

// 

// Example: 

// "dog", "bird", "elephant", "dog", "dog", "cat" 

// "cat", "dog", "dog", "cat", "cat", "fish" 

// Result (order doesn't matter) 

// "dog", "dog", "cat"

*/

Solution

#include <iostream>
#include <list>
#include <iterator>
#include <algorithm>
#include <string>

using namespace std;

void find_comm_strings(list<string>& output, list<string>& listA, list<string>& listB)
{
    listA.sort();
    listB.sort();

    list<string>::const_iterator citA = listA.begin();
    list<string>::const_iterator citB = listB.begin();

    while(citA != listA.end() && citB != listB.end()){
        int eq = (*citA).compare(*citB);
        if(eq == 0){
            output.push_back(*citA);
            citA ++;
            citB ++;
        }
        else if (eq > 0){
            citB ++;
        }
        else{
            citA ++;
        }
    }
}

int main(int argc, char* argv[])
{
    list<string> listA;
    list<string> listB;
    list<string> output;

    cout << "list A:" << endl;
    listA.push_back("dog");
    listA.push_back("bird");
    listA.push_back("elephant");
    listA.push_back("dog");
    listA.push_back("dog");
    listA.push_back("cat");
    copy(listA.begin(), listA.end(), ostream_iterator<string>(cout, ","));
    cout << endl;

    cout << "list B:" << endl;
    listB.push_back("cat");
    listB.push_back("dog");
    listB.push_back("dog");
    listB.push_back("cat");
    listB.push_back("cat");
    listB.push_back("fish");
    copy(listB.begin(), listB.end(), ostream_iterator<string>(cout, ","));
    cout << endl;

    find_comm_strings(output, listA, listB);

    cout << "common strings" << endl;
    copy(output.begin(), output.end(), ostream_iterator<string>(cout, ","));
    cout << endl;

 return 0;
}

Output

list A:
dog,bird,elephant,dog,dog,cat,
list B:
cat,dog,dog,cat,cat,fish,
common strings
cat,dog,dog,
Press any key to continue . . .

文章结束给大家分享下程序员的一些笑话语录： 问：你觉得让你女朋友（或者任何一个女的）从你和李彦宏之间选一个，你觉得她会选谁？　　
　　答：因为李艳红这种败类，所以我没女友！