OJ练习44——T5 Longest Palindormic Substring

求字符串的最长回文子串。

【思路】

1.从两边开始向中间发展

2.从中间开始向两边发展

3.从中间开始的变体，较为复杂，详见http://articles.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html

【other code1-思路2】

 string longestPalindrome(string s) {
        int n=s.size();
        if(n==0) return "";
        string re=s.substr(0,1);
        for(int i=0; i<n; i++){
            string p1=testPalindrome(s,i,i);
            if(p1.size()>re.size())
                re=p1;
            string p2=testPalindrome(s,i,i+1);
            if(p2.size()>re.size())
                re=p2;
        }
        return re;
    }
    string testPalindrome(string s, int c1, int c2){
        int l=c1;
        int r=c2;
        int n=s.size();
        while(l>=0&&r<=n-1&&s[l]==s[r]){
            l--;
            r++;
        }
        return s.substr(l+1,r-l-1);
    }

【结果1】

时间是O(n^2),100+ms，排名靠后。

【other code2-思路3】

// 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);
  int n = T.length();
  int *P = new int[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 maxLen = 0;
  int centerIndex = 0;
  for (int i = 1; i < n-1; i++) {
    if (P[i] > maxLen) {
      maxLen = P[i];
      centerIndex = i;
    }
  }
  delete[] P;
  
  return s.substr((centerIndex - 1 - maxLen)/2, maxLen);
}

【结果2】

时间是O(n), 12ms，排名第一 

为毛中档的题目就这么难！  (ง •̀_•́)ง┻━┻

下午分析。