17. Algorithm Exercises: Manacher Algorithm

Manacher algorithm core

1) Understand the palindrome radius array

2) Understand the rightmost boundary R of the palindrome of all centers, and the center point C when obtaining R

3) Understand the structure of L...(i`)...C...(i)...R, and the division of situations based on the length of i' palindrome

4) Each situation can be divided to speed up the process of solving the i palindrome radius.

Exercise 1 Assume that the length of the string str is N. If you want to return the length of the longest palindrome substring, the time complexity is O(N).

public static int manacher(String s) {
		if (s == null || s.length() == 0) {
			return 0;
		}
		// "12132" -> "#1#2#1#3#2#"
		char[] str = manacherString(s);
		// 回文半径的大小
		int[] pArr = new int[str.length];
		int C = -1;
		// 讲述中：R代表最右的扩成功的位置
		// coding：最右的扩成功位置的，再下一个位置
		int R = -1;
		int max = Integer.MIN_VALUE;
		for (int i = 0; i < str.length; i++) { // 0 1 2
			// R第一个违规的位置，i>= R
			// i位置扩出来的答案，i位置扩的区域，至少是多大。
			pArr[i] = R > i ? Math.min(pArr[2 * C - i], R - i) : 1;
			while (i + pArr[i] < str.length && i - pArr[i] > -1) {
				if (str[i + pArr[i]] == str[i - pArr[i]])
					pArr[i]++;
				else {
					break;
				}
			}
			if (i + pArr[i] > R) {
				R = i + pArr[i];
				C = i;
			}
			max = Math.max(max, pArr[i]);
		}
		return max - 1;
	}

	public static char[] manacherString(String str) {
		char[] charArr = str.toCharArray();
		char[] res = new char[str.length() * 2 + 1];
		int index = 0;
		for (int i = 0; i != res.length; i++) {
			res[i] = (i & 1) == 0 ? '#' : charArr[index++];
		}
		return res;
	}

	// for test
	public static int right(String s) {
		if (s == null || s.length() == 0) {
			return 0;
		}
		char[] str = manacherString(s);
		int max = 0;
		for (int i = 0; i < str.length; i++) {
			int L = i - 1;
			int R = i + 1;
			while (L >= 0 && R < str.length && str[L] == str[R]) {
				L--;
				R++;
			}
			max = Math.max(max, R - L - 1);
		}
		return max / 2;
	}

	// for test
	public static String getRandomString(int possibilities, int size) {
		char[] ans = new char[(int) (Math.random() * size) + 1];
		for (int i = 0; i < ans.length; i++) {
			ans[i] = (char) ((int) (Math.random() * possibilities) + 'a');
		}
		return String.valueOf(ans);
	}

	public static void main(String[] args) {
		int possibilities = 5;
		int strSize = 20;
		int testTimes = 5000000;
		System.out.println("test begin");
		for (int i = 0; i < testTimes; i++) {
			String str = getRandomString(possibilities, strSize);
			if (manacher(str) != right(str)) {
				System.out.println("Oops!");
			}
		}
		System.out.println("test finish");
	}

Exercise 2: Given the head nodes head1 and head2 of two binary trees, you want to know whether there is a subtree in head1 that has the same structure as head2.

public static String shortestEnd(String s) {
		if (s == null || s.length() == 0) {
			return null;
		}
		char[] str = manacherString(s);
		int[] pArr = new int[str.length];
		int C = -1;
		int R = -1;
		int maxContainsEnd = -1;
		for (int i = 0; i != str.length; i++) {
			pArr[i] = R > i ? Math.min(pArr[2 * C - i], R - i) : 1;
			while (i + pArr[i] < str.length && i - pArr[i] > -1) {
				if (str[i + pArr[i]] == str[i - pArr[i]])
					pArr[i]++;
				else {
					break;
				}
			}
			if (i + pArr[i] > R) {
				R = i + pArr[i];
				C = i;
			}
			if (R == str.length) {
				maxContainsEnd = pArr[i];
				break;
			}
		}
		char[] res = new char[s.length() - maxContainsEnd + 1];
		for (int i = 0; i < res.length; i++) {
			res[res.length - 1 - i] = str[i * 2 + 1];
		}
		return String.valueOf(res);
	}

	public static char[] manacherString(String str) {
		char[] charArr = str.toCharArray();
		char[] res = new char[str.length() * 2 + 1];
		int index = 0;
		for (int i = 0; i != res.length; i++) {
			res[i] = (i & 1) == 0 ? '#' : charArr[index++];
		}
		return res;
	}

	public static void main(String[] args) {
		String str1 = "abcd123321";
		System.out.println(shortestEnd(str1));
	}