Shortest Palindrome

最新推荐文章于 2022-11-01 19:17:22 发布

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最新推荐文章于 2022-11-01 19:17:22 发布

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本文介绍了一种通过在字符串前添加字符将其转换为最短回文字符串的方法，并提供了一个具体的Java实现方案。该方法利用Manacher算法计算字符串的最长回文子串，从而高效地构造出所需的最短回文字符串。

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Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

public class Solution {
    public String shortestPalindrome(String s) {
        if (s == null || s.length() == 0) {
        	return "";
        }
        String copy = s;
        StringBuffer stringBuffer = new StringBuffer(s);
        StringBuffer reverse = stringBuffer.reverse();
        s = reverse.toString();
        char[] charArr = manacherString(s);
        int[] pArr = new int[charArr.length];
        int index = -1;
        int pR = -1;
        int max = -1;
        for (int i = 0; i < pArr.length; i++) {
        	pArr[i] = pR > i ? Math.min(pArr[2 * index - i], pR - i) : 1;
        	while (i + pArr[i] < charArr.length && i - pArr[i] > -1) {
        		if (charArr[i + pArr[i]] == charArr[i - pArr[i]]) {
        			pArr[i]++;
        		} else {
        			break;
        		}
        	}
        	if (i + pArr[i] > pR) {
        		pR = i + pArr[i];
        		index = i;
        	}
        	if (pR == charArr.length) {
        		max = pArr[i];
        		break;
        	}
		}
        char[] tmp = new char[s.length() - max + 1];
        for (int i = 0; i < tmp.length; i++) {
        	tmp[tmp.length - 1 - i] = charArr[i * 2 + 1];
		}
        return new StringBuffer(new String(tmp)).reverse() + copy;
    }

	private char[] manacherString(String s) {
		// TODO Auto-generated method stub
		char[] charArray = s.toCharArray();
		char[] res = new char[s.length() * 2 + 1];
		int index = 0;
		for (int i = 0; i < res.length; i++) {
			res[i] = (i & 1) == 0 ? '#' : charArray[index++];
		}
		return res;
	}
}