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：

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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;    }}}

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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);    }};