5. Longest Palindromic Substring（unsolved）

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”

解答：这道题还没看懂，得再看

class Solution {
public:
    string longestPalindrome(string s) {
    if (s.empty()) return "";
    if (s.size() == 1) return s;
    int min_start = 0, max_len = 1;
    for (int i = 0; i < s.size();) {
      if (s.size() - i <= max_len / 2) break;
      int j = i, k = i;
      while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters.
      i = k+1;
      while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand.
      int new_len = k - j + 1;
      if (new_len > max_len) { min_start = j; max_len = new_len; }
    }
    return s.substr(min_start, max_len);
}
};

二刷时的做法：

class Solution {
public:
    string longestPalindrome(string s) {
        if(s.empty()) return s;
        int left=0,right=0,index=0,len=0;
        for(int i=0;i<s.size();i++){
            if(s[i]==s[i+1]){
               left=i;
               right=i+1;
               search(s,left,right,index,len);
            }
            left=i;
            right=i;
            search(s,left,right,index,len);
        }
        if(len==0) len=s.size();
        return s.substr(index,len);

    }
    void search(string& s,int left,int right,int &index,int &len){
        int step=1;
        while((left-step)>=0&&(right+step)<s.size()){
            if(s[left-step]!=s[right+step]) break;
            step++;
        }
        int wide=right-left+2*step-1;
        if(len<wide){
            len=wide;
            index=left-step+1;
        }

    }
};