5. Longest Palindromic Substring

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.

思路：(beat 2.63%)

　　以一个位置为准，考察两侧相同距离是否相同。这样会导致"aaaaaaaaaaaaaaaaaaaaaaaaaa..."达到最坏O(n²)的时间复杂度。

　　而且这种情况下需要对两种回文类型分开提取。

代码 C++：

class Solution {
public:
    string longestPalindrome(string s) {
        if (s.length() <= 1)
            return s;
        string solve = "";

        for (int flag = 0; flag < s.length() - solve.length() / 2; flag++) {
            string temp = "";
            temp.push_back(s[flag]);
            for (int i = flag - 1,j = flag + 1; i >=0 && j < s.length() && s[i] == s[j]; i--,j++) {
                string ins = "";
                ins.push_back(s[i]);
                temp.insert(0, ins);
                temp.push_back(s[i]);
            }
            if (temp.length() > solve.length())
                solve = temp;
        }
        for (int flag = 0; flag < s.length() - solve.length() / 2; flag++) {
            string temp = "";
            for (int i = flag,j = flag + 1; i >=0 && j < s.length() && s[i] == s[j]; i--,j++) {
                string ins = "";
                ins.push_back(s[i]);
                temp.insert(0, ins);
                temp.push_back(s[i]);
            }
            if (temp.length() > solve.length())
                solve = temp;
        }
        return solve;
    }
};

Reference: https://leetcode.com/discuss/32204/simple-c-solution-8ms-13-lines

思路：

用i遍历字符串，j,k初始化与i相等，k++ if 相等，这样，j,k就把中奖相同的给框住了，然后再j--k++找到相同，用min_pos记录回文串首位置，max_length记录回文串长度，即j-k+1。

Θ(1)存储空间，将两种回文串一同解决，且没有上述O(n²)的最坏情况

代码：

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);
}