5. Longest Palindromic Substring

最新推荐文章于 2025-07-16 16:59:19 发布

原创最新推荐文章于 2025-07-16 16:59:19 发布 · 384 阅读

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本文探讨了如何在给定字符串中找到最长的回文子串。通过递归方法和动态规划方法进行了实现，并对存储结构进行了优化以提高效率。

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.
原题

这题开始想用递归，没写出来，看了题解，原来可以用动态规划，但是用vector存储的时候会超时，需要优化下存储结构。

不太懂下面这种用递归做的方法？

#include <algorithm>
#include <iostream>
#include <map>
#include <string.h>
#include <string>
#include <vector>
using namespace std;

string find_palimdrom(string s, int left, int right) {
  int n = s.size();
  int l = left;
  int r = right;
  while (l >= 0 && r <= n - 1 && s[l] == s[r]) {
    l--;
    r++;
  }
  return s.substr(l + 1, r - l - 1);
}

string palimdrom_recursive_way(string s) {
  if (s.size() <= 1)
    return s;
  string longest, str;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    str = find_palimdrom(s, i, i);
    if (str.size() > longest.size())
      longest = str;
    str = find_palimdrom(s, i, i + 1);
    if (str.size() > longest.size())
      longest = str;
  }
  return longest;
}

//但是使用vector在leetcode上会超时
string palimdrom_DP_way(string s) {
  int n = s.size();
  if (n <= 1)
    return s;
  vector<vector<bool>> matrix(n, vector<bool>(n));
  string longest;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j < n; j++) {
      if (i == j || (j - i < 2 && s[i] == s[j]) ||
          (s[i] == s[j] && matrix[i + 1][j - 1])) {
        matrix[i][j] = true;

        if (longest.size() < j - i + 1)
          longest = s.substr(i, j - i + 1);
      }
    }
  }
  return longest;
}

//优化版：优化的存储方式
string palimdrom_DP_OP_way(string s) {
  // Construct a matrix, and consdier matrix[j][i] as s[i] -> s[j] is Palindrome
  // or not.
  //                                 ------^^^^^^
  //                                 NOTE: it's [j][i] not [i][j]
  bool **matrix = new bool *[n];
  int start, end;
  for (int i = 0; i < n; i++) {
    matrix[i] = new bool[i + 1];
    memset(matrix[i], false, (i + 1) * sizeof(bool));
    matrix[i][i] = true;
    for (int j = 0; j < i; j++) {
      if (i == j || (s[j] == s[i] && (i - j < 2 || matrix[i - 1][j + 1]))) {
        matrix[i][j] = true;
        if (len < i - j + 1) {
          start = j;
          len = i - j + 1;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    delete[] matrix[i];
  }
  delete[] matrix;

  return s.substr(start, len);
}

int main() {
  string s;
  cin >> s;
  // cout << palimdrom_recursive_way(s) << endl;

  // cout << palimdrom_DP_way(s).size() << endl;
  cout << palimdrom_recursive_way(s).size() << endl;
}