1050. String Subtraction (20)

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本文介绍了一种高效计算两个字符串差集的方法，通过利用映射查询表优化时间复杂度至O(N)，并提供了详细代码实现。适用于需要快速处理字符串匹配与去除任务的场景。

Given two strings S₁ and S₂, S = S₁ - S₂ is defined to be the remaining string after taking all the characters in S₂ from S₁. Your task is simply to calculate S₁ - S₂ for any given strings. However, it might not be that simple to do it fast.

Input Specification:

Each input file contains one test case. Each case consists of two lines which gives S₁ and S₂, respectively. The string lengths of both strings are no more than 10⁴. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.

Output Specification:

For each test case, print S₁ - S₂ in one line.

Sample Input:

They are students.
aeiou

Sample Output:

Thy r stdnts.

注意点：这道题能过的时间复杂度应该是O(N) 设计好算法就OK

参考代码：

#include <iostream>
#include <vector>
#include <string>
#include <map>
using namespace std;
map<char,vector<int> > myMap;

int main()
{
    string s;
    getline(cin,s);
    for(int i=0;i<s.length();i++){
        myMap[s[i]].push_back(i);
    }
    string chk;
    getline(cin,chk);
    for(int i=0;i<chk.length();i++){
        if(!myMap[chk[i]].empty()){
            myMap[chk[i]].clear();
        }
    }
    string last[10001];
    map<char,vector<int> >::iterator it;
    for(it=myMap.begin();it!=myMap.end();it++){
        if(!(*it).second.empty()){
            char ch = (*it).first;
            for(int i=0;i<(*it).second.size();i++){
                last[(*it).second[i]] = ch;
            }
        }
    }
    for(int i=0;i<10001;i++){
        if(last[i]!=""){
            cout<<last[i];
        }
    }
    return 0;
}

利用ASCLL来建立映射查询表，代码更简单

参考AC代码：

#include<iostream>
#include <vector>
#include <string>
using namespace std;
#define  MAX 1000
int main()
{
	string a, b;
	getline(cin, a);
	getline(cin, b);
	vector<bool> visited(MAX, false);
	for (int i = 0; i < b.size(); ++i)
		visited[b[i]] = true;       //利用char类型的ASCLL码来建立映射查询表
	for (int i = 0; i < a.size(); ++i)
		if (!visited[a[i]])
            cout<<a[i];
	return 0;
}