PAT1050 String Subtraction (20分) 用数组即可

最新推荐文章于 2021-08-04 22:59:54 发布

wwang_dev

最新推荐文章于 2021-08-04 22:59:54 发布

阅读量184

点赞数

CC 4.0 BY-SA版权

分类专栏： c/c++ algorithm PAT 文章标签：字符串 c++

本文链接：https://blog.youkuaiyun.com/wwang_dev/article/details/107112973

algorithm 同时被 3 个专栏收录

56 篇文章

订阅专栏

c/c++

54 篇文章

订阅专栏

PAT

52 篇文章

订阅专栏

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.

解析

所谓 s1 - s2 就是从s1中去除出现在s2中的字符，输出剩余部分。
所以可以逐个判断s1的字符，判断其是否在s2中出现，若未出现则输出，否则跳过即可。
为了操作简单，使用一个布尔数组flag[256]，ascii字符也就是0-255，所以遍历一次s2，将flag[s2[i]]设置为true即表示当前字符在s2中出现过。

代码

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

int main() {
    string s1, s2;
    // 原字符串中包含空格，使用getline
    getline(cin, s1);
    getline(cin, s2);
    int len1 = s1.length(), len2 = s2.length();
    bool flag[256] = {false};
    // 在s2中出现的字符，在s1中剔除
    for (int i = 0; i < len2; ++i) flag[s2[i]] = true;
    // 逐个字符判断，若未在s2中出现则输出
    for (int i = 0; i < len1; ++ i)
        if (!flag[s1[i]]) cout << s1[i];
    return 0;
}