字符串循环移位

标签：c++   字符串移位   循环移位   

第一种方法为把右边的向左移，每次移动n位。一直到末尾，再把末尾的反转好

//字符串左移n位
void shift(std::string &str, int n)
{
	int len = str.length();
	if (len <= 0) return;
	n = n%len;

	int start = 0;//交换起始位置
	int end = n;//后面交换的起始位置
	int mark = end;//记录这个位置
	while (true)
	{
		while (start != mark&&end < len)
		{
			char temp = str[start];
			str[start] = str[end];
			str[end] = temp;
			++start;
			++end;
			//std::cout << str << std::endl;
		}
		if (start == mark)
		{
			if (end >= len)
				return;
			else
				mark = end;//更新mark
		}
		else if (end >= len)
			end = mark;
	}

}

//反转[start end]之间字符串
void inversion(std::string& str, int start, int end)
{
	char temp;
	while (start<end)
	{
		temp = str[start];
		str[start] = str[end];
		str[end] = temp;
		++start;
		--end;
	}
}

void shift1(std::string& str, int n)
{
	int len = str.length();
	if (len <= 0) return;
	n = n%len;
	inversion(str, 0, n - 1);//反转前n半部分
	inversion(str, n, len - 1);//反转后半部分
	inversion(str, 0, len - 1);//整体反转
}

//计算公约数
int gcd(int max, int min)
{
	if (min == 0)
		return max;
	return gcd(min, max%min);
}
void shift2(std::string& str, int n)
{
	int len = str.length();
	if (len <= 0) return;
	n = n%len;
	int GCD = gcd(len, n);
	int loop = len / GCD;

	for (int i = 0; i < GCD; ++i)//循环最大公约数次
	{
		char temp = str[i];
		int j;
		for (j = 0; j < loop - 1; ++j)//每次循环换loop个。
		{
			str[(i + j*n) % len] = str[(i + j*n + n) %len ];
		}
		str[(i + j*n) % len] = temp;
	}
}

int main()
{
	std::string str = "abcde";
	shift(str, 3);
	std::cout << str << std::endl;
	shift1(str, 3);
	std::cout << str << std::endl;
	shift2(str, 3);
	std::cout << str << std::endl;
	return 0;
}

字符串循环移位

标签：c++   字符串移位   循环移位   

原文地址：http://blog.csdn.net/kangroger/article/details/40511729