Leetcode5[Med] 最长回文字串-DP
思路借鉴:DP，下面是 Python代码.
class Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
matrix = [[0 for i in range(len(s))] for i in range(len(s))] //initialize matrix to keep track of the state
result = "" //record longest sequence
max_len = 0 //record longest length
for j in range(len(s)):
for i in range(0, j+1):
if j - i <=1: //i=j or i,j 相邻 即考虑:a或bb情况
if s[i] == s[j]:
matrix[i][j] = True
if max_len < j-i+1:
result = s[i:j+1]
max_len = j - i + 1
else: //考虑abba情况
if s[i] == s[j] and matrix[i+1][j-1]:
matrix[i][j] = True
if max_len < j-i+1:
result = s[i:j+1]
max_len = j - i + 1
return result
下面是 Java代码.
class Solution {
public String longestPalindrome(String s) {
int len = s.length();
boolean[][] dp = new boolean[len][len];
String result = "";
int max_len = 0;
for(int j=0; j<len; j++){
for(int i=0; i<=j; i++){
if (j-i <=1){
if (s.charAt(i) == s.charAt(j)){
dp[i][j] = true;
if(max_len<j-i+1){
result = s.substring(i,j+1);
max_len = j - i + 1;
}
}
}
else{
if (s.charAt(i)==s.charAt(j) && dp[i+1][j-1]){
dp[i][j] = true;
if(max_len<j-i+1){
result = s.substring(i,j+1);
max_len = j - i + 1;
}
}
}
}
}
return result;
}
}