ZOJ Problem Set - 2478 Encoding

最新推荐文章于 2019-09-23 21:41:27 发布

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最新推荐文章于 2019-09-23 21:41:27 发布

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分类专栏：基础算法-----字符和字符串

本文链接：https://blog.youkuaiyun.com/Wchenchen0/article/details/81777790

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本文介绍了一种简单的字符串压缩编码算法，该算法通过识别并压缩连续重复的字符来减少字符串的长度。文章提供了两种实现方法，一种使用了错误的方法（通过排序和映射），另一种则是正确的实现方式。正确的方法通过遍历字符串并计数重复字符的数量来进行压缩。

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Given a string containing only 'A' - 'Z', we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. If the length of the sub-string is 1, '1' should be ignored.

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 100.

Output

For each test case, output the encoded string in a line.

Sample Input

2
ABC
ABBCCC

Sample Output

ABC
A2B3C

这题一开始我理解错了意思，以为是把所有的字母从小到大排好序来算的，用map错了

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
int main()
{
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--) {
		string s;
		s.clear();
		cin >> s;
		map<char, int>mp;
		mp.clear();
		for (int i = 0;s[i];i++) {
			mp[s[i]]++;
		}
		map<char, int>::iterator it;
		for (it = mp.begin();it != mp.end();++it) {
			if (it->second == 1)cout << it->first;
			else cout << it->second << it->first;
		}
		cout << endl;
	}
	return 0;
}

AC代码

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<iostream>
using namespace std;
char s[10005];
int main()
{
	int T;
	cin>>T;
	while (T--) {
		int flag = 1;
		cin>>s;
		int len = strlen(s);
		for (int i = 0;i<len;i++)
		{
			if (s[i] == s[i + 1])
				flag++;
			else {
				if (flag>1)
					cout<<flag;
				cout<<s[i];
				flag = 1;
			}

		}
		cout << endl;;
	}

	return 0;
}