LeetCode 3.5 Longest Palindromic Substring

3.5 Longest Palindromic Substring

描述

Given a stringS, find the longest palindromic substring inS. You may assume that the maximumlength ofS is 1000, and there exists one unique longest palindromic substring.

3.5 Longest Palindromic Substring61

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(n2)。思路二:记忆化搜索,复杂度O(n2)。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(n2)。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


true , i= j

f(i, j) =S[i] = S[j], j =i + 1S[i]=S[j]andf(i+1,j−1),j>i+1

思路三: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

// LeetCode, 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)typedef string::const_iterator Iterator;
  namespace std {
  template<>
  struct hash<pair<Iterator, Iterator>> {
      size_t operator()(pair<Iterator, Iterator> const& p) const {
          return ((size_t) &(*p.first)) ^ ((size_t) &(*p.second));
}};
}
  class Solution {
  public:
      string longestPalindrome(string const& s) {
          cache.clear();
          return cachedLongestPalindrome(s.begin(), s.end());
      }
  private:
      unordered_map<pair<Iterator, Iterator>, string> cache;
      string longestPalindrome(Iterator first, Iterator last) {
          size_t length = distance(first, last);
          if (length < 2) return string(first, last);
          auto s = cachedLongestPalindrome(next(first), prev(last));
          if (s.length() == length - 2 && *first == *prev(last))
              return string(first, last);
          auto s1 = cachedLongestPalindrome(next(first), last);
          auto s2 = cachedLongestPalindrome(first, prev(last));
          // return max(s, s1, s2)
          if (s.size() > s1.size()) return s.size() > s2.size() ? s : s2;
          else return s1.size() > s2.size() ? s1 : s2;
}
      string cachedLongestPalindrome(Iterator first, Iterator last) {
          auto key = make_pair(first, last);
          auto pos = cache.find(key);
          if (pos != cache.end()) return pos->second;
          else return cache[key] = longestPalindrome(first, last);
      }
};

动规

// LeetCode, 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(string s) {
const int n = s.size();
bool f[n][n];
fill_n(&f[0][0], n * n, false);
// 用vector会超时
//vector<vector<bool> > f(n, vector<bool>(n, false));size_t max_len = 1, start = 0; //最长回文子串的长度,起点
          for (size_t i = 0; i < s.size(); i++) {
              f[i][i] = true;
for(size_tj=0;j<i;j++){ //[j,i]
f[j][i] = (s[j] == s[i] && (i - j < 2 || f[j + 1][i - 1]));if (f[j][i] && max_len < (i - j + 1)) {
                      max_len = i - j + 1;
start = j;}
}}
          return s.substr(start, max_len);
      }
};

Manaer’s Algorithm

// LeetCode, Longest Palindromic Substring// Manacher’s Algorithm
// 时间复杂度O(n),空间复杂度O(n)
class Solution {
  public:
      // 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
      string preProcess(string s) {
          int n = s.length();
          if (n == 0) return "^$";
          string ret = "^";
          for (int i = 0; i < n; i++) ret += "#" + s.substr(i, 1);
ret += "#$";
return ret;
}
string longestPalindrome(string s) {
string T = preProcess(s);
const 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[n];
          int C = 0, R = 0;
          for (int i = 1; i < n - 1; i++) {
              int i_mirror = 2 * C - i; // equals to i' = C - (i-C)
              P[i] = (R > i) ? min(R - i, P[i_mirror]) : 0;
              // Attempt to expand palindrome centered at i
              while (T[i + 1 + P[i]] == T[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 max_len = 0;
          int center_index = 0;
          for (int i = 1; i < n - 1; i++) {
              if (P[i] > max_len) {
                  max_len = P[i];
                  center_index = i;
}}
    return s.substr((center_index - 1 - max_len) / 2, max_len);
}
};