跳到主要内容

Longest Palindromic Substring

描述​

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

分析​

最长回文子串，非常经典的题。

思路一：暴力枚举，以每个元素为中间元素，同时从左右出发，复杂度O(n^2)。

思路二：记忆化搜索，复杂度O(n^2)。设f[i][j] 表示[i,j]之间的最长回文子串，递推方程如下：

f[i][j] = if (i == j) S[i]
          if (S[i] == S[j] && f[i+1][j-1] == S[i+1][j-1]) S[i][j]
          else max(f[i+1][j-1], f[i][j-1], f[i+1][j])

思路三：动规，复杂度O(n^2)。设状态为f(i,j)，表示区间[i,j]是否为回文串，则状态转移方程为

f(i,j)={true,i=jS[i]=S[j],j=i+1S[i]=S[j] and f(i+1,j−1),j>i+1f(i,j)=\begin{cases} true & ,i=j\\ S[i]=S[j] & , j = i + 1 \\ S[i]=S[j] \text{ and } f(i+1, j-1) & , j > i + 1 \end{cases}

思路四：Manacher’s Algorithm, 复杂度O(n)。详细解释见 http://leetcode.com/2011/11/longest-palindromic-substring-part-ii.html。

备忘录法​

// Longest Palindromic Substring
// 备忘录法，会超时
// 时间复杂度O(n^2)，空间复杂度O(n^2)
public class Solution {
    private final HashMap<Pair, String> cache = new HashMap<>();

    public String longestPalindrome(final String s) {
        cache.clear();
        return cachedLongestPalindrome(s, 0, s.length() - 1);
    }

    String longestPalindrome(final String s, int i, int j) {
        final int length = j - i + 1;
        if (length < 2) return s.substring(i, j + 1);

        final String s1 = cachedLongestPalindrome(s, i + 1, j - 1);

        if (s1.length() == length - 2 && s.charAt(i + 1) == s.charAt(j - 1))
            return s.substring(i, j + 1);

        final String s2 = cachedLongestPalindrome(s, i + 1, j);
        final String s3 = cachedLongestPalindrome(s, i, j - 1);

        // return max(s1, s2, s3)
        if (s1.length() > s2.length()) return s1.length() > s3.length() ? s1 : s3;
        else return s2.length() > s3.length() ? s2 : s3;
    }

    String cachedLongestPalindrome(final String s, int i, int j) {
        final Pair key = new Pair(i, j);

        if (cache.containsKey(key)) {
            return cache.get(key);
        } else {
            final String result = longestPalindrome(s, i, j);
            cache.put(key, result);
            return result;
        }
    }

    // immutable
    static class Pair {
        private int x;
        private int y;

        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public int hashCode() {
            return x * 31 + y;
        }

        @Override
        public boolean equals(Object other) {
            if (this == other) return true;
            if (this.hashCode() != other.hashCode()) return false;
            if (!(other instanceof Pair)) return false;

            final Pair o = (Pair) other;
            return this.x == o.x && this.y == o.y;
        }
    }
}

动规​

// Longest Palindromic Substring
// 动规，时间复杂度O(n^2)，空间复杂度O(n^2)
class Solution {
    public String longestPalindrome(final String s) {
        final int n = s.length();
        final boolean[][] f = new boolean[n][n];
        int maxLen = 1, start = 0;  // 最长回文子串的长度，起点

        for (int i = 0; i < n; i++) {
            f[i][i] = true;
            for (int j = 0; j < i; j++) {  // [j, i]
                f[j][i] = (s.charAt(j) == s.charAt(i) &&
                        (i - j < 2 || f[j + 1][i - 1]));
                if (f[j][i] && maxLen < (i - j + 1)) {
                    maxLen = i - j + 1;
                    start = j;
                }
            }
        }
        return s.substring(start, start + maxLen);
    }
}

Manacher’s Algorithm​

// Longest Palindromic Substring
// Manacher’s Algorithm
// 时间复杂度O(n)，空间复杂度O(n)
class Solution {
    // Transform S into T.
    // For example, S = "abba", T = "^#a#b#b#a#$".
    // ^ and $ signs are sentinels appended to each end to avoid bounds checking
    public String preProcess(final String s) {
        int n = s.length();
        if (n == 0) return "^$";

        StringBuilder ret = new StringBuilder("^");
        for (int i = 0; i < n; i++) ret.append("#" + s.charAt(i));

        ret.append("#$");
        return ret.toString();
    }

    String longestPalindrome(String s) {
        String T = preProcess(s);
        final int n = T.length();
        // 以T[i]为中心，向左/右扩张的长度，不包含T[i]自己，
        // 因此 P[i]是源字符串中回文串的长度
        int[] P = new int[n];
        int C = 0, R = 0;

        for (int i = 1; i < n - 1; i++) {
            int iMirror = 2 * C - i; // equals to i' = C - (i-C)

            P[i] = (R > i) ? Math.min(R - i, P[iMirror]) : 0;

            // Attempt to expand palindrome centered at i
            while (T.charAt(i + 1 + P[i]) == T.charAt(i - 1 - P[i]))
                P[i]++;

            // If palindrome centered at i expand past R,
            // adjust center based on expanded palindrome.
            if (i + P[i] > R) {
                C = i;
                R = i + P[i];
            }
        }

        // Find the maximum element in P.
        int maxLen = 0;
        int centerIndex = 0;
        for (int i = 1; i < n - 1; i++) {
            if (P[i] > maxLen) {
                maxLen = P[i];
                centerIndex = i;
            }
        }

        final int start =(centerIndex - 1 - maxLen) / 2;
        return s.substring(start, start + maxLen);
    }
}