LeetCode5 longest palindromic substring（最长回文子串）

题目

Given a string s, return the longest palindromic substring in s. 
 Example 1: 
Input: s = "babad"
Output: "bab"
Note: "aba" is also a valid answer.

 Example 2: 
Input: s = "cbbd"
Output: "bb"

 Example 3: 
Input: s = "a"
Output: "a"

 Example 4: 
Input: s = "ac"
Output: "a"

 Constraints: 
 1 <= s.length <= 1000 
 s consist of only digits and English letters. 

方法

Manacher法

1.首先要将每个字符的间隔加上分隔符#，以此统一奇数和偶数
2.遍历字符串，记录确认为回文的最右边界maxR,并且用p[i]存储第i个字符的最长回文半径，假设当前i小于maxR,i的对称位置是j，如果i+p[j]和maxR的最小值作为判断回文的起点进行优化

• 时间复杂度：O(n)
• 空间复杂度：O(n)

class Solution {
    public String longestPalindrome(String s) {
        String s1 = preProcess(s);
        int length = s1.length();
        int c=0,maxR=0,maxLength=0,start=0;
        int[] p = new int[length];
        for(int i=1;i<length-1;i++){
            if(i<maxR){
                p[i] = Math.min(maxR-i,p[2*c-i]);
            }
            int left = i-p[i]-1;
            int right = i+p[i]+1;
            while(left>=0&&right<length&&s1.charAt(left)==s1.charAt(right)){
                left--;
                right++;
                p[i]++;
            }
            if(maxR<p[i]+i){
                maxR = p[i]+i;
                c = i;
            }
            if(maxLength<p[i]){
                maxLength = p[i];
                start = (i-p[i])/2;
            }
        }
        return s.substring(start,start+maxLength);
    }
    private String preProcess(String s){
        StringBuffer strBuf = new StringBuffer("^#");
        for(int i=0;i<s.length();i++){
            strBuf.append(s.charAt(i)+"#");
        }
        strBuf.append("$");
        return strBuf.toString();
    }
}