5. Longest Palindromic Substring

最新推荐文章于 2024-01-20 11:35:01 发布

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最新推荐文章于 2024-01-20 11:35:01 发布

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本文探讨了两种高效算法——中心扩展法和马拉车算法，用于寻找字符串中的最长回文子串。通过实例解析，详细解释了算法原理及其实现过程。

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Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

solution（核心是从左到右选中一个点，或者两个相邻点向两侧扩充，代码中避免区分是一点，两点的方法值得借鉴）

其中马拉车法

class Solution {
    
   private int l,r;
    public String longestPalindrome(String s) {
        if(s.length()<2){
            return s;
        }
        l=r=0;
        int len = s.length();
        for(int i=0;i<len;i++){
            extendPalindromic(s,i,i);
            extendPalindromic(s,i,i+1);
        }
        return  s.substring(l,r+1);
    }

    private void extendPalindromic(String s,int left,int right){
        while(left>=0&&right<s.length()&&s.charAt(left)==s.charAt(right)){
            left--;
            right++;
        }
        
        if(r-l<=right-left-2){
            r=right-1;
            l=left +1;
        }
    }
}

方法2： Manacher's Algorithm（核心是对以往的中心点向两侧的最大对称距离值加以利用，降低了检索时间；并利用插入#的方法避免了类似“aa”偶数对称）感觉与KMP算法类似。

public String longestPalindrome(String s) {
        /* Preprocess s: insert '#' between characters, so we don't need to worry about even or odd length palindromes. */
        char[] newStr = new char[s.length() * 2 + 1];
        newStr[0] = '#';
        for (int i = 0; i < s.length(); i++) {
            newStr[2 * i + 1] = s.charAt(i);
            newStr[2 * i + 2] = '#';
        }
        /* Process newStr */
        /* dp[i] is the length of LPS centered at i */
        int[] dp = new int[newStr.length];
        /**
         * For better understanding, here we define "friend substring", or "friend":
         * "friend substring" has the largest end-index in all checked substrings that
         * are palindromes. We start at friendCenter = 0 and update it in each cycles.
         */
        int friendCenter = 0, friendRadius = 0, lpsCenter = 0, lpsRadius = 0;
        /* j is the symmetry of i with respect to friendCenter */
        int j;
        for (int i = 0; i < newStr.length; i++) {
            /* Calculate dp[i] */
            if (friendCenter + friendRadius > i) {
                /**
                 * This is the most important part of the algorithm.
                 *
                 * Normally we start from dp[i] = 1 and then try to expand dp[i] by doing brute-force palindromic
                 * checks. However, if i is in the range of friend (friendCenter + friendRadius > i), we can expect
                 * dp[i] = dp[j] because friend is a palindrome. This only works within the range of friend, so the
                 * max value of dp[i] we can trust is (friendEnd - i).
                 *
                 * Here is an example:
                 *
                 *     friendStart   j             friendCenter  i     friendEnd
                 *               |   |             |             |     |
                 * String: - - d c b a b c d - - - - - - - d c b a b c ? - - - - - - - -
                 *               [--------friend (palindrome)--------]
                 *
                 * In this example, (friendEnd - i) = 3, so we can only be certain that radius <= 3 part around i
                 * is a palindrome (i.e. "cbabc" part). We still need to check the character at "?".
                 */
                j = friendCenter - (i - friendCenter);
                dp[i] = Math.min(dp[j], (friendCenter + friendRadius) - i);
            }
            else {
                /* Calculate from scratch */
                dp[i] = 1;
            }
            /* Check palindrome and expand dp[i] */
            while (i + dp[i] < newStr.length && i - dp[i] >= 0&& newStr[i + dp[i]] == newStr[i - dp[i]])
                dp[i]++;
            /* Check if i should become the new friend */
            if (friendCenter + friendRadius < i + dp[i]) {
                friendCenter = i;
                friendRadius = dp[i];
            }
            /* Update longest palindrome */
            if (lpsRadius < dp[i]) {
                lpsCenter = i;
                lpsRadius = dp[i];
            }
        }
        return s.substring((lpsCenter - lpsRadius + 1) / 2, (lpsCenter + lpsRadius - 1) / 2);
    }

参考地址：https://leetcode.com/problems/longest-palindromic-substring/discuss/128054/Java-(Beats-98)-and-JavaScript-(Beats-100)-O(n)-Manacher's-Algorithm-with-Detailed-Explanation