public String longestPalindrome(String s) { int maxLength = 0; int maxStart = 0; int len = s.length(); //i是字符串长度 for(int i = 0; i < len; i++){ //j是字符串起始位置 for(int j = 0; j < len - i; j++){ //挨个判断是否回文 if(
*/publicstaticStringlongestPalindrome(String s){if(s==null||s.length()<2){returns;}int maxLength=0;String longest=null;int length=s.length();boolean[][]table=newboolean[length][length];// 单个字符都是回文for(int i=0;i<length;i++){table[i][i]=true;longest=s.substring(i,i+1);ma...
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...
Can you solve this real interview question? Longest Palindromic Substring - Given a string s, return the longest palindromic substring in s. Example 1: Input: s = "babad" Output: "bab" Explanation: "aba" is also a valid answer. Example 2: Input
LeetCode the longest palindrome substring 回文检测,参考http://blog.csdn.net/feliciafay/article/details/16984031 使用时间复杂度和空间复杂度相对较低的动态规划法来检测,具体的做法 图一 偶数个回文字符情况 图二 奇数个回文字符情况 核心就是如果一个子串是回文,如果分别向回文左右侧扩展一个字符相同,那么回文...
对于字符串S, 要找到它最长的回文子串,能想到的最暴力方法,应该是对于每个元素i-th都向左向右对称搜索,最后用一个数组span 记录下相对应元素i-th为中心的回文子串长度。 那么问题来了: 1. 这样的方法,对于奇回文子串和偶回文子串的处理不一样,比如所“acbca” 和“acbbca” ...
Below image shows the output of the above longest palindrome java program. We can improve the above code by moving the palindrome and longest lengths check into a different function. However, I have left that part for you. :) Please let me know if there are any other better implementations...
java 代码 publicclassSolution{publicStringlongestPalindrome(Strings){if(s==null||s.length()<=1)returns;char[]arr=s.toCharArray();intlen=arr.length;intstart=0;intlength=1;booleanpali[][]=newboolean[len][len];for(inti=0;i<pali.length;i++){pali[i][i]=true;// length = 1;}for(inti...
java暴力publicclasslongestPalindrome{publicstaticvoidmain(String[]args){Stringres=longestPalindrome("babad");System.out.println(res);//String s = "asdedfg";}publicstaticStringlongestPalindrome(Strings){Stringresult="";Stringtmp;for(inti=0;i=i;j--){tmp=s.substring(i,j+1);if(getRoll(tmp)&&...
1classSolution {2intlo, maxLen;34publicString longestPalindrome(String s) {5if(s ==null|| s.length() == 0)return"";67for(inti = 0; i < s.length(); i ++) {8findPalindrome(s, i, i);//assume odd length, try to extend Palindrome as possible9findPalindrome(s, i, i + 1);/...