PAT 1050. String Subtraction (20)

最新推荐文章于 2021-08-21 17:07:52 发布

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本文介绍了一种高效算法，用于从一个字符串中移除另一个字符串的所有字符。通过使用类似于哈希表的数据结构来标记待删除字符，实现了快速查找和处理。此方法避免了逐个字符比对可能带来的性能瓶颈。

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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.

思路：题目的要求是在s1字符串中消除掉s2字符串里面所有的字符，用find直接判定s1的每个字符是否在s2中可能会超时。考虑到ASCII值得范围，我们可以利用类似hash的判定数组来判定。

 #include <iostream>
#include <string>

using namespace std;

int main()
{

	int a[200] = {0};
	string s1, s2;
	getline(cin, s1);
	getline(cin, s2);
	for (int i=0; i!=s2.size(); ++i) //要删除的字符的ASCII值设为1
	           a[s2[i]] = 1;
	for (int i=0; i!=s1.size(); ++i)
	    if (!a[s1[i]])    //当字符不是要删除的才输出
	      cout << s1[i];
	cout << endl;
	//system("pause");
	return 0;
}