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问题描述
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.
解决思路
1. 动态规划:时间空间复杂度均为O(n^2);
2. 中心扩展法:注意考虑奇数长度和偶数长度的情况,时间复杂度为O(n^2);
3. Manacher算法:
(1) 用#夹杂在字符串中,进行填充,首字符为$;
(2) 动态规划求解,记录最长回文子串的长度和中心位置;
(3) 去除多余的字符;
时间空间复杂度均为O(n).
程序
// dp public class Solution { public String longestPalindrome(String s) { if (s == null || s.length() == 0) { return ""; } int n = s.length(); int begin = 0, maxLen = 1; boolean[][] isPalindrome = new boolean[n][n]; // 初始化 // 单个字符为回文串 for (int i = 0; i < n; i++) { isPalindrome[i][i] = true; } // 连续两个相同的字符也是回文串 for (int i = 0; i < n - 1; i++) { if (s.charAt(i) == s.charAt(i + 1)) { isPalindrome[i][i + 1] = true; begin = i; maxLen = 2; } } for (int i = n -3; i >= 0; i--) { for (int j = i + 2; j < n; j++) { if (s.charAt(i) == s.charAt(j) && isPalindrome[i + 1][j - 1]) { isPalindrome[i][j] = true; if (j - i + 1 > maxLen) { maxLen = j - i + 1; begin = i; } } } } return s.substring(begin, begin + maxLen); } }
public class LongestPalindromicSubstring { // 中心匹配法, O(n^2), O(1) public String longestPalindrome(String s) { if (s == null || s.length() == 0) { return ""; } int len = s.length(); String res = ""; for (int i = 0; i < len; i++) { String s1 = longestPString(s, i, i); String s2 = longestPString(s, i, i + 1); String longer = s1.length() > s2.length() ? s1 : s2; if (longer.length() > res.length()) { res = longer; } } return res; } private String longestPString(String s, int i, int j) { int low = 0, high = s.length() - 1; while (i >= low && j <= high) { if (s.charAt(i) != s.charAt(j)) { break; } --i; ++j; } return s.substring(i + 1, j); } // Manacher算法 public String longestPalindrome2(String s) { if (s == null || s.length() == 0) { return ""; } // fill $ and # String filled = filledString(s); int len = filled.length(); int[] r = new int[len]; r[0] = 1; int center = 0, maxR = 0; for (int i = 1; i < len; i++) { int maxLeft = center + maxR; if (i < maxLeft) { r[i] = Math.min(r[2 * center - i], maxLeft - i); } else { r[i] = 1; } // max expand while (i - r[i] >= 0 && i + r[i] < len && filled.charAt(i - r[i]) == filled.charAt(i + r[i])) { ++r[i]; } if (r[i] > maxR) { maxR = r[i]; center = i; } } return removeSign(filled.substring(center - maxR + 1, center + maxR)); } private String removeSign(String s) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c == ‘#‘) { continue; } sb.append(c); } return sb.toString(); } private String filledString(String s) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < s.length(); i++) { sb.append("#"); sb.append(s.charAt(i)); } sb.append("#"); return "$" + sb.toString(); } }
Manacher算法图解
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原文地址:http://www.cnblogs.com/harrygogo/p/4679580.html