Mannacher算法

查找最長迴文子串用暴力解的話時間複雜度 O(n^2) ，利用 Mannacher 算法實現時間複雜度爲 O(n) 。其思想主要是利用迴文結構的對稱性來解決，類似於用表的方式紀錄信息，避免重複計算。

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

int getLen(string s2, int n)
{
	int len = 0;
	while (n - len >= 0 && n + len <= s2.length())
	{
		if (s2[n - len] != s2[n + len])
		{
			break;
		}
		len++;
	}
	return len;
}

int main()
{
	string s1 = "abcdcbdbcdcbdbcd";
	string s2 = "#";
	for (int i = 0; i < s1.length(); i++)
	{
		s2 += s1[i];
		s2 += "#";
	}
	int* count = new int[s2.length()];
	int i = 0;
	int right = -1;
	int mid = -1;
	while (i < s2.length())
	{
		if (i > right)  // 第一種情況
		{
			count[i] = getLen(s2, i);
			right += count[i];
			mid = i;
		}
		else
		{
			if (count[2 * mid - i] + i -1 > right)  // 第三種情況
			{
				count[i] = right - i;
			}
			else if (count[2 * mid - i] + i -1 == right)  // 第四種情況
			{
				count[i] = getLen(s2, i);
				mid = i;
			}
			else
			{
				count[i] = count[2 * mid - i];  // 第二種情況
			}
		}
		i++;
	}
	cout << "暴力求解:" << endl;
	for (int i = 0; i < s2.length(); i++)
	{
		cout << getLen(s2, i) << " ";
	}
	cout << endl;
	cout << "Mannacher算法:" << endl;
	for (int i = 0; i < s2.length(); i++)
	{
		cout << count[i] << " ";
	}
	cout << endl;

	system("pause");
	return 0;
}