leetcode Longest Palindromic Substring Part II-优快云博客

本文链接：https://blog.youkuaiyun.com/taoqick/article/details/14169261

本文介绍了一种高效算法来查找给定字符串中的最长回文子串，通过利用回文的对称特性实现O(n)的时间复杂度。文章详细解释了算法的工作原理，并提供了具体的代码实现。

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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, and there exists one unique longest palindromic substring.

There is little to say about the O(n^2) solution.

The O(n) solution is by virtue of symmetric property.

if P[ i' ] ≤ R – i,
then P[ i ] ← P[ i' ]
else P[ i ] ≥ P[ i' ]. (Which we have to expand past the right edge (R) to find P[ i ].

If the palindrome centered at i does expand past R, we update C to i, (the center of this new palindrome), and extend R to the new palindrome’s right edge.

So the code is :

class Solution {
 public:
  string parse(string& s) {
    string res = "^#";
    for (int i = 0; i < s.length(); ++i)
      res += s.substr(i,1) + "#";
    res += "$";
    return res;
  }
  string longestPalindrome(string s) {
    if (s.length() <= 1)
      return s;
    string T = parse(s);
    int n = T.length();
    vector<int> P(n,0);
    int i, j, C = 0, R = 0, ii;
    for (i = 1; i < n - 1; ++i) {
      ii = C*2 - i;
      P[i] = (R - i) > 0 ? min(P[ii], R - i) : 0;
      while (T[i - 1 - P[i]] == T[i + 1 + P[i]])
        ++P[i];
      if (i + P[i] > R){
        R = i + P[i];
        C= i;    
      }
    }
    int max = 0;
    int maxIndex = 1;
    for (i = 1; i < n - 1; ++i) {
      if (P[i] > max) {
        max = P[i];
        maxIndex = i;
      }
    }
    return s.substr( (maxIndex - max - 1 ) / 2,max);
  }
};

Add some explanations where the complexity is O(n), for the step if (i+P[i] > R):

1. If i+P[i] <= R, the while loop will execute only once because of the sysmetric property

2. If i+P[i]>R, P[i] starts from R-i. The updates for R is linear.