PAT甲级 1050. String Subtraction (20)

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;
}


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