问题 D: String Subtraction (20)（哈希）

最新推荐文章于 2022-08-15 23:31:26 发布

原创最新推荐文章于 2022-08-15 23:31:26 发布 · 256 阅读

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本文介绍了一个字符串操作问题的解决方案，即计算两个给定字符串S1和S2的差集S1-S2，通过使用哈希映射记录字符出现情况，高效地找出并输出所有在S1中但不在S2中的字符。

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 Copy

They are students.
aeiou

样例输出 Copy

Thy r stdnts.

#include<bits/stdc++.h>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int main(int argc, char** argv) {
	int hash_map[200] = {0};
	string s1, s2;
	while(getline(cin, s1)) {
		getline(cin, s2);
		int len1 = s1.length();
		int len2 = s2.length();
		for(int i = 0; i < len1; i++) {//把出现过的字符标记1
			hash_map[s1[i]] = 1;
		}
		for(int i = 0; i < len2; i++) {//需要去掉的字符标记0
			hash_map[s2[i]] = 0;
		}
		for(int i = 0; i < len1; i++) {
			if(hash_map[s1[i]]){//标记1的字符才输出
				cout << s1[i];
			}
		}
		cout << endl;
	}
	return 0;
}