《算法笔记》4.2小节——算法初步-＞哈希问题 D: String Subtraction (20)-优快云博客

该程序实现了一个简单的字符串减法操作，即从一个字符串中移除另一个字符串的所有字符。它使用了C++编程语言，通过创建一个字符集合来存储第二个字符串中的所有字符，然后遍历第一个字符串，跳过存在于集合中的字符，输出剩余的字符。这个方法的时间复杂度为O(n)，其中n为较长字符串的长度。

问题 D: String Subtraction (20)

时间限制 : 1.000 sec 内存限制 : 32 MB

题目描述

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 - S2for any given strings. However, it might not be that simple to do it fast.

输入

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.

输出

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

样例输入

They are students.
aeiou

样例输出

Thy r stdnts.

#include <iostream>
#include <set>
#include <cstring>

using namespace std;


int main() {
    char s1[10005], s2[10005];
    while (gets(s1)) {
        gets(s2);
        int len_s1 = strlen(s1);
        int len_s2 = strlen(s2);
        set<char> set;
        for (int i = 0; i < len_s2; ++i) {
            set.insert(s2[i]);
        }
        for (int i = 0; i < len_s1; ++i) {
            if (set.count(s1[i]) == 1)
                continue;
            else
                cout << s1[i];
        }
        cout << endl;
    }
    return 0;
}