String Subtraction (20)

最新推荐文章于 2024-05-17 06:07:48 发布

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最新推荐文章于 2024-05-17 06:07:48 发布

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分类专栏： codeup 文章标签：哈希字符串减法

本文链接：https://blog.youkuaiyun.com/AliceGoToAnother/article/details/79363093

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本文介绍了一种计算两个字符串差集的算法实现，通过构建哈希表来高效地去除第二个字符串中包含的所有字符，最终输出第一个字符串中未被第二个字符串覆盖的部分。

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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.

#include <cstdio>
#include <cstring>
//将字符转换成整数
int charToNum(char a)
{
    int num = 0;
    if(a >= 'A' && a <= 'Z') num = a - 'A';
    else if(a >= 'a' && a <= 'z') num = a - 'a' + 26;
    return num;
}

int main()
{
    char str[10010];
    char del[10010];
    gets(str);
    gets(del);
    int len1 = strlen(str);
    int len2 = strlen(del);
    int hash[53] = {0};
    for(int i = 0;i < len1;i++)
    {
        if(str[i] != ' ' && str[i] != '.')
            hash[charToNum(str[i])] = 1;
    }
    for(int i = 0;i < len2;i++)
    {
        if(hash[charToNum(del[i])] == 1)
        {
            hash[charToNum(del[i])] = 0;
        }
    }
    for(int i = 0;i < len1;i++)
    {
        if(hash[charToNum(str[i])] == 1 || str[i] == ' ' || str[i] == '.') printf("%c",str[i]);
    }
    printf("\n");
    return 0;
}