PAT甲级 1050. String Subtraction (20)

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本文介绍了一种高效的字符串处理方法，通过统计目标字符串中字符的出现频率，实现快速从源字符串中移除这些字符的功能。该方法适用于可见ASCII码组成的字符串，通过对输入字符串进行预处理，可以显著提高处理速度。

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题目：

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：

1.根据常规做法，在s1中查找s2的字符并删除，有两个测试点运行超时；

2.利用题目中输入字符均为可见ASCII码和空格，先利用数组统计s2中字符出现情况。在对s1进行处理时，直接利用数组判断当前字符是否需要删除。这里要注意，删除后，角标要-1，因为长度会缩短。

代码：

#include<iostream>
#include<string>
using namespace std;

int main()
{
	ifstream cin;
	cin.open("case1.txt");
	string s1, s2;
	getline(cin,s1);
	getline(cin,s2);

	int char2num[127] = { 0 };
	int i, j,k;	
	for (i = 0; i < s2.size(); ++i)
	{
		k = (int)s2[i];
		char2num[k]++;
	}
	
	for (j = 0; j < s1.size(); ++j)
	{
		k = (int)s1[j];
		if (char2num[k])
		{
			s1.erase(j,1);
			--j;
		}
	}

	cout << s1 << endl;

	system("pause");
	return 0;
}