回文判断

来源：互联网发布：计算机应用与软件期刊编辑：程序博客网时间：2024/06/11 23:27

算法1：暴力解法，枚举所有子串，对每个子串判断是否为回文，复杂度为O(n^3)

import java.util.Scanner;/** * 最长回文子串，暴力解决，找到字符串中的每一个子串，记录下来其中最长的 * @author buder_cp * */public class 最长回文子串暴力 {public static boolean isPanlidrome(String s, int start, int end) {for (int i = start, j = end; i < j; i++, j--) {if (s.charAt(i) != s.charAt(j)) {return false;}}return true;}public static String longestPalidrome(String s) {int n = s.length();int max = 0;int from = 0;int to = 1;for (int i = 0; i < n; i++) {for (int j = i; j < n; j++) {if (isPanlidrome(s, i, j)) {if (j - i + 1 > max) {max = j - i + 1;from = i;to = j;}}}}return s.substring(from, to + 1);}public static void main(String[] args){Scanner sc = new Scanner(System.in);System.out.println(longestPalidrome(sc.nextLine()));}}

算法2：以某个元素为中心，分别计算偶数长度的回文最大长度和奇数长度的回文最大长度。时间复杂度O(n^2)，空间O（1）

public class 最长回文子串中心扩展 {public static void main(String[] args){String s = "12321";int len = s.length();int max = 1;int start = 0;for(int i = 1; i < len; i++){int low = i-1;//偶数情况int high = i;while(low >= 0 && high < len && s.charAt(low)==s.charAt(high)){low--;high++;}if(high-low-1 > max){max = high-low-1;start = low+1;}low = i-1;//奇数情况high = i+1;while(low >= 0 && high < len && s.charAt(low)==s.charAt(high)){low--;high++;}if(high-low-1 > max){max = high-low-1;start = low+1;}}System.out.println(max);}}

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