【算法笔记4.2小节-哈希】问题 D: String Subtraction (20)（字符哈希）

本文链接：https://blog.youkuaiyun.com/xunalove/article/details/87904801

本文介绍了一种计算两个字符串S1和S2相减的方法，即从S1中去除所有S2中存在的字符。通过使用哈希表记录S2中字符的出现情况，再遍历S1进行对比，实现快速计算。

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题目链接：http://codeup.cn/problem.php?cid=100000582&pid=3

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

They are students.
aeiou

样例输出

Thy r stdnts.

思路

将字符的ASCII码作为字符的唯一标示。直接用int转换即可。


#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    char a[10010], b[10010];
    gets(a);
    gets(b);
    int hashTable[128] = {0};//ASCII为7为编码最多为127
    for(int i=0; i<strlen(b); i++)
    {
        hashTable[(int)b[i]]=1;
    }
    int k=0;
    char ans[10010];
    for(int i=0; i<strlen(a); i++)
    {
        if(hashTable[(int)a[i]]==0)
            ans[k++] = a[i];
    }
    ans[k++]='\0';		//不要忘记
    puts(ans);
    return 0;

}