Problem Statement |
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| The Happy Letter game is played as follows: At the beginning, several players enter the field. Each player has a lowercase English letter on their back. The game is played in turns. In each turn, you select two players with different letters, and both selected players leave the field. The game ends once it is impossible to take another turn. If there are some players left in the field at the end of the game, they must all have the same letter. That letter is called the winning letter. If there are no players left in the field at the end of the game, there is no winning letter. You are given a string letters. The characters in letters are the characters carried by the players at the beginning of the game. Return a string with all possible winning letters. The letters in the returned string must be sorted in increasing order. |
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Definition |
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Limits |
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Notes |
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| - | If there's no happy letter, return the empty string. | ||||||||||||
Constraints |
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| - | letters will contain between 1 and 50 elements. | ||||||||||||
| - | Each element of letters will be a lowercase English letter ('a'-'z'). | ||||||||||||
Examples |
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Mean:
给你一个只有小写字母组成的字符串，每一轮你可以选择两个不相同的字符删去，如果最后还有剩下的字符，那么这个字符就是winning letter，现在要你返回可能是winning letter的字符组成的字符串，并按照升序排序。
analyse:
我们首先将每个字母出现的次数统计一遍，然后就是对26个小写字母进行判断，如果不包括本身，最多数量的那个字符的数量大于剩余字符的数量，说明不可能满足题目的要求(不同的字符相互匹配)，否则就符合题目要求，加入ans字符串即可。
Time complexity：O(n)
Source code：
// BEGIN CUT HERE
// END CUT HERE
#line 5 "HappyLetterDiv1.cpp"
//Memory Time
// K MS
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1100
#define LL long long
using namespace std;
class HappyLetterDiv1 {
public:
string getHappyLetters(string letters) {
string ans;
ans.clear();
int len=letters.size();
int cnt[30]={0};
for(int i=0;i<len;i++)
cnt[letters[i]-'a']++;
for(int i=0;i<26;i++)
{
if(cnt[i])
{
if(cnt[i]==1&&!(len&1))
continue;
int maxx=-1;
for(int j=0;j<26;j++)
if(j!=i)
maxx=max(maxx,cnt[j]);
if(maxx<=len-1-maxx)
ans+='a'+i;
}
}
return ans;
}
};
