【PAT】A1050 String Subtraction (20分)--字符(串)hash

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本文介绍了一种使用散列思想简化字符串减法运算的方法，通过建立散列表标记已出现字符，实现对两个字符串进行快速减法操作。

1050 String Subtraction (20分)

-----题目地址-----
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 S1and S2 , respectively. The string lengths of both strings are no more than 10^4 . 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.

作者 CHEN, Yue 单位浙江大学

思路总结：

直接对字符串做删除操作显然有点麻烦，实际可以用散列的思想使问题得到简化。
①做散列表bool H[200] ，初始状态均为false。
②遍历字符串2 : str，对字符串2中的每个字符str[i]，令H[str[i]-'\0'] = true (直接写H[str[i]] = true也没问题)，表示该字符出现过。
③遍历字符串1: all，对all中每个字符，若该字符在散列表H中未出现也即false，则输出，反之不输出。

AC代码:

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

string all, str;
bool H[200] = {false};
int main()
{
    getline(cin, all);
    getline(cin,str);

    for(int i=0; i<str.length(); i++){
        int key = str[i] - '\0';
        H[key] = true;
    }
    for(int i=0; i<all.length(); i++){
        int key = all[i] - '\0';
        if(H[key] == false){
            printf("%c",all[i]);
        }
    }
    return 0;
}