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LongestPalindromicSubstring.java
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/*
Author: King, wangjingui@outlook.com
Date: Dec 13, 2014
Problem: Longest Palindromic Substring
Difficulty: Medium
Source: https://oj.leetcode.com/problems/longest-palindromic-substring/
Notes:
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.
Solution: 1. Time O(n^2), Space O(n^2)
2. Time O(n^2), Space O(n)
3. Time O(n^2), Space O(1) (actually much more efficient than 1 & 2)
4. Time O(n), Space O(n) (Manacher's Algorithm)
5. Time O(n), Smaller Space than solution 4. (Manacher's Algorithm)
*/
public class Solution {
public String longestPalindrome_1(String s) {
int n = s.length();
boolean[][] dp = new boolean[n][n];
int idx = 0, maxLen = 0;
for (int k = 0; k < n; ++k) {
for (int i = 0; i + k < n; ++i) {
if (k == 0 || k == 1) dp[i][i+k] = (s.charAt(i) == s.charAt(i+k));
else dp[i][i+k] = (s.charAt(i) == s.charAt(i+k)) ? dp[i+1][i+k-1] : false;
if (dp[i][i+k] == true && (k+1) > maxLen) {
idx = i; maxLen = k + 1;
}
}
}
return s.substring(idx, idx + maxLen);
}
public String longestPalindrome_2(String s) {
int n = s.length();
boolean[][] dp = new boolean[2][n];
int idx = 0, maxLen = 0;
int cur = 1, last = 0;
for (int i = 0; i < n; ++i) {
cur = cur + last - (last = cur);
for (int j = i; j >=0; --j) {
if (j == i || j == i - 1)
dp[cur][j] = (s.charAt(i) == s.charAt(j));
else dp[cur][j] = (s.charAt(i) == s.charAt(j)) && dp[last][j + 1];
if (dp[cur][j] && (i - j + 1) > maxLen) {
idx = j; maxLen = i - j + 1;
}
}
}
return s.substring(idx, idx + maxLen);
}
public String longestPalindrome_3(String s) {
int n = s.length();
int idx = 0, maxLen = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j <= 1; ++j) {
boolean isP = true;
for (int k = 0; i - k >= 0 && i + j + k < n && isP; ++k) {
isP = (s.charAt(i - k) == s.charAt(i + j + k));
if (isP && (j + 1 + k*2) > maxLen) {
idx = i - k; maxLen = j + 1 + k*2;
}
}
}
}
return s.substring(idx, idx + maxLen);
}
public String longestPalindrome_4(String s) {
int n = s.length();
int idx = 0, maxLen = 0;
StringBuffer sb = new StringBuffer();
sb.append('^');
for (int i = 0; i < n; ++i) {
sb.append('#');
sb.append(s.charAt(i));
}
sb.append("#$");
n = 2 * n + 3;
int mx = 0, id = 0;
int[] p = new int[n];
Arrays.fill(p,0);
for (int i = 1; i < n - 1; ++i) {
p[i] = (mx > i) ? Math.min(p[2 * id - i], mx - i) : 0;
while (sb.charAt(i + 1 + p[i]) == sb.charAt(i - 1 - p[i])) ++p[i];
if (i + p[i] > mx) {
id = i; mx = i + p[i];
}
if (p[i] > maxLen) {
idx = i; maxLen = p[i];
}
}
idx = (idx - maxLen - 1) / 2;
return s.substring(idx, idx + maxLen);
}
public String longestPalindrome_5(String s) {
int n = s.length();
int idx = 0, maxLen = 0;
int mx = 0, id = 0;
int[] p = new int[2*n+1];
Arrays.fill(p,0);
for (int i = 0; i < 2*n+1; ++i) {
p[i] = (mx > i) ? Math.min(p[2*id-i], mx - i) : 0;
int left = i - 1 - p[i], right = i + 1 + p[i];
while (left>=0 && right <= 2*n) {
if (left % 2 == 0 || s.charAt(left/2) == s.charAt(right/2)) {
++p[i];
} else break;
--left;
++right;
}
if (i + p[i] > mx) {
id = i; mx = i + p[i];
}
if (p[i] > maxLen) {
idx = i; maxLen = p[i];
}
}
idx = (idx - maxLen) / 2;
return s.substring(idx, idx + maxLen);
}
}