leetcode214. shortest palindromic sequence

214. shortest palindrome string

Difficulty 114

Given a string  S , you can add a character string in the front to convert it to a palindromic sequence. Find and return the shortest in this manner may be converted palindromic sequence.

Example 1:

Input: "aacecaaa"
Output: "aaacecaaa"

Example 2:

Input: "abcd"
Output: "dcbabcd"

Thinking: horse-drawn vehicles starting with 0 determined maximum palindrome sequence, flipped and then add back to the front of the string.

Fool can understand a horse-drawn carriage Manacher

class Solution {
public String preProcess(String s) {
    int n = s.length();
    if (n == 0) {
        return "^$";
    }
    String ret = "^";
    for (int i = 0; i < n; i++)
        ret += "#" + s.charAt(i);
    ret += "#$";
    return ret;
}

// 马拉车算法
public String shortestPalindrome(String s) {
    String T = preProcess(s);
    int n = T.length();
    int[] P = new int[n];
    int C = 0, R = 0;
    for (int i = 1; i < n - 1; i++) {
        int i_mirror = 2 * C - i;
        if (R > i) {
            P[i] = Math.min(R - i, P[i_mirror]);// 防止超出 R
        } else {
            P[i] = 0;// 等于 R 的情况
        }

        // 碰到之前讲的三种情况时候，需要利用中心扩展法
        while (T.charAt(i + 1 + P[i]) == T.charAt(i - 1 - P[i])) {
            P[i]++;
        }

        // 判断是否需要更新 R
        if (i + P[i] > R) {
            C = i;
            R = i + P[i];
        }

    }

    //这里的话需要修改
    int maxLen = 0;
    int centerIndex = 0;
    for (int i = 1; i < n - 1; i++) {
        int start = (i - P[i]) / 2;
        //我们要判断当前回文串是不是开头是不是从 0 开始的
        if (start == 0) {
            maxLen = P[i] > maxLen ? P[i] : maxLen;
        }
    }
    return new StringBuilder(s.substring(maxLen)).reverse() + s;
}
}