5. Longest Palindromic Substring

题目

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

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

解析

• 想到了左右拓展的方法，但是开始不知道怎么处理
• 动态规划的方法

class Solution_5 {
public:

    // reverse(res.begin(),res.end());
    // return s == string(s.rbegin(), s.rend());
    bool isPalindrome(string s)
    {
        return s ==string(s.rbegin(),s.rend());
    }

    // 处理bb和aba的情况
    string expandstring(const string s,int l,int r)
    {
        while(l >= 0&&r<s.size()&&s[l]==s[r])
        {
            l--;
            r++;
        }

        return s.substr(l+1,r-l-1);
    }

    string longestPalindrome(string s) {
        if (s.size()<=1)
        {
            return s;
        }   
        string longestr = s.substr(0, 1);
        for (int i = 0; i < s.size()-1;i++)
        {
            string expand = expandstring(s, i, i);
            if (expand.length()>longestr.size())
            {
                longestr = expand;
            }
            string expand2 = expandstring(s,i,i+1);
            if (expand2.length()>longestr.length())
            {
                longestr = expand2;
            }
        }
        return longestr;
    }
};

string longestPalindromeDP(string s) {
  int n = s.length();
  int longestBegin = 0;
  int maxLen = 1;
  bool table[1000][1000] = {false};
  for (int i = 0; i < n; i++) {
    table[i][i] = true;
  }
  for (int i = 0; i < n-1; i++) {
    if (s[i] == s[i+1]) {
      table[i][i+1] = true;
      longestBegin = i;
      maxLen = 2;
    }
  }
  for (int len = 3; len <= n; len++) {   //对每个长度拓展查看
    for (int i = 0; i < n-len+1; i++) {
      int j = i+len-1;
      if (s[i] == s[j] && table[i+1][j-1]) {
        table[i][j] = true;
        longestBegin = i;
        maxLen = len;
      }
    }
  }
  return s.substr(longestBegin, maxLen);
}

题目来源

• 5. Longest Palindromic Substring
• Longest Palindromic Substring Part I