字符串之判断重复字符串

1.去掉重复字符串时间复杂度为O（n）

#include <iostream>
using namespace std;

int main()
{
	char str[] = "bdsjiseftftftfyrzsesese";
	int length = strlen(str); 
	char *p = str;
	char hashTable[256] = {0};
	
	for (int i  = 0;i <length;++i)
	{
		if (hashTable[str[i]]==0)
		{
			*p = str[i];
			hashTable[*p] =1;
			p++;
		}

	}
	*p ='\0';
	cout << str <<endl;
	return 0;
}

 2.第一次只出现一次的字符

#include <iostream>
using namespace std;

int main()
{
	char *str = "edsjiseftftftfyrzsesese";
	
	char *p = str;
	char hashTable[256] = {0};
	
	while(*p)
	{
		hashTable[*p]++;
		p++;
	}
	p = str;
	while(*p)
	{
		if(hashTable[*p] == 1)
		{
			break;
		}
		p++;
	}
	cout << *p <<endl;
	return 0;
}

 

3.删除出现次数最少的字符

有三个关键字 次数 最少 删除

需要遍历三次：

void reverse(char *str)
{
	char hashtable[26] = {0};
	char *p = str;
	while (*p != '\0')
	{
		hashtable[*p -'a']++;
		p++;
	}
	p = str;
	int min = hashtable[*p -'a'];
	while(*p != '\0')
	{
		if (hashtable[*p - 'a'] != 0)
		{
			if(hashtable[*p - 'a'] < min)
			{
				min = hashtable[*p -'a'];
			}
		}
		p++;
	}
	p = str;
	for (int i  =  0 ;i < strlen(str);i++)
	{
		if (hashtable[str[i] - 'a'] != min)
		{
			*p = str[i];
			p++;
		}
	}
}