摘要:是左闭右开区间,所以要。,要理解是和之间只有一个元素。循环每次的时候,都要更新子串更大的情况。补一种中点延展的方法循环字符串的每个字符,以该字符为中心,若两边为回文,则向两边继续延展。循环返回长度最长的回文串即可。
Problem
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.
ExampleGiven the string = "abcdzdcab", return "cdzdc".
ChallengeO(n2) time is acceptable. Can you do it in O(n) time.
Note动规好题。
.substring(start, end)是左闭右开区间,所以end要+1。
if (s.charAt(i) == s.charAt(j) && (i - j <= 2 || dp[j+1][i-1])),要理解i - j <= 2是i和j之间只有一个元素。
循环每次i - j > end - start的时候,都要更新子串更大的情况。Time和space都是O(n^2)。
补一种中点延展的方法:循环字符串的每个字符,以该字符为中心,若两边为回文,则向两边继续延展。如此,每个字符必对应一个最长回文串。循环返回长度最长的回文串即可。
Solution DP</>复制代码
public class Solution {
public String longestPalindrome(String s) {
// Write your code here
int len = s.length();
boolean[][] dp = new boolean[len][len];
int start = 0, end = 0;
for (int i = 0; i < len; i++) {
for (int j = 0; j <= i; j++) {
if (s.charAt(i) == s.charAt(j) && (i - j <= 2 || dp[j+1][i-1])) {
dp[j][i] = true;
if (end - start < i - j ) {
start = j;
end = i;
}
}
}
}
return s.substring(start, end+1);
}
}
方法二:中点延展
</>复制代码
public class Solution {
String longest = "";
public String longestPalindrome(String s) {
for (int i = 0; i < s.length(); i++) {
helper(s, i, 0);
helper(s, i, 1);
}
return longest;
}
public void helper(String s, int i, int os) {
int left = i, right = i + os;
while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
left--;
right++;
}
String cur = s.substring(left+1, right);
if (cur.length() > longest.length()) {
longest = cur;
}
}
}
2018-02-03 Added some comments, same thoughts as solution 2
</>复制代码
class Solution {
public String longestPalindrome(String s) {
if (s == null || s.length() < 2) return s;
//return as result
String longest = s.substring(0, 1);
for (int i = 0; i < s.length()-1; i++) {
//get "ABA" type palindrome
String cur = getPalindrome(s, i, i);
//get "ABBA" type palindrome, and compare its length with "ABA" type
if (s.charAt(i+1)==s.charAt(i)) {
String temp = getPalindrome(s, i, i+1);
cur = cur.length() > temp.length() ? cur : temp;
}
//update longest with cur
longest = longest.length() > cur.length() ? longest : cur;
}
return longest;
}
public String getPalindrome(String s, int left, int right) {
while (left > 0 && right < s.length()-1 && s.charAt(left-1) == s.charAt(right+1)) {
left--;
right++;
}
//careful with the right bound
return s.substring(left, right+1);
}
}
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