Leetcode 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”

思路：

从中间向两边扫描

代码：

---

Java

public class Solution {
private int lo, maxLen;

public String longestPalindrome(String s) {
    int len = s.length();
    if (len < 2)
        return s;

    for (int i = 0; i < len-1; i++) {
         extendPalindrome(s, i, i);  //assume odd length, try to extend Palindrome as possible
         extendPalindrome(s, i, i+1); //assume even length.
    }
    return s.substring(lo, lo + maxLen);
}

private void extendPalindrome(String s, int j, int k) {
    while (j >= 0 && k < s.length() && s.charAt(j) == s.charAt(k)) {
        j--;
        k++;
    }
    if (maxLen < k - j - 1) {
        lo = j + 1;
        maxLen = k - j - 1;
    }
}}

---

DP

public String longestPalindrome(String s) {
    boolean[][] flag = new boolean[s.length()][s.length()];
    int maxlen = 0,start = 0;
    for(int i = 0;i < s.length(); i++){
        flag[i][i] = true;
        maxlen = 1;
        start = i;
    }
    for(int i = 0;i < s.length()-1; i++)
        if(s.charAt(i)==s.charAt(i+1)){
            flag[i][i+1] = true;
            maxlen = 2;
            start = i;
        }
    for(int len = 3; len<= s.length(); len++)
        for(int i = 0;i < s.length()-len+1; i++){
            int j = i+len-1;
            if(s.charAt(i)==s.charAt(j)&&flag[i+1][j-1]==true){
                flag[i][j] = true;
                maxlen = len;
                start = i;
            }
        }
    return s.substring(start, start+maxlen);
}

c++

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);
    }
};