importjava.util.Scanner;publicclassPalindrome{publicstaticvoidmain(String[] args){Scannersc=newScanner(System.in);longn;while((n = sc.nextLong()) !=0){// 输入以零结束longnum;booleanbook=true;Stringstr=null;for(inti=2;i <=16;i++) {// (2-16) 进制num = decimalToMRadix(n, i);// ...
); } // length of the string // if there is one character string - returing true int len = str.length(); if (len <= 1) { return true; } // Converting the string into uppercase // to make the comparisons case insensitive String strU = str.toUpperCase(); // result variable /...
法I:逐一比较最左端与最右端的数字 classSolution {publicbooleanisPalindrome(intx) {if(x<0)returnfalse;intleft;//the first bit of the inteterintright;//the last bit of the integerintdivisor = 1;//divisor of each iterationinttmp =x;//determine the initial value of divisorwhile(x/divisor ...
System.out.println(longestPalindromeString("1223213")); System.out.println(longestPalindromeString("abb")); } static public String intermediatePalindrome(String s, int left, int right) { if (left > right) return null; while (left >= 0 && right < s.length() && s.charAt(left) == s.ch...
int i = 0, j = s.length() - 1; while(i < j) { if(s.charAt(i) != s.charAt(j)) return false; i++; j--; } return true; } } 这个博客还用了一种set的方法: http://www.cnblogs.com/grandy... 暂时还没看懂,明天再看下。。
//每次都比较x最高位和最低位,不一样了就说明不是回文数 //如果到最后都一样说明x是回文数 func isPalindrome(x int) bool { if x < 0 { return false } oX := x //high表示x的数量级 high := 1 for oX >= 10 { high *= 10 oX /= 10 } for x > 0 { //left为最高位 left := ...
int max = 1; String res = String.valueOf(s.charAt(0)); for (int i = 1; i < dp.length; i++) { for (int j = 0; j <= s.length() - i; j++) { if (i == 1) dp[i][j] = true; //one char is always palindrome else if ( i == 2) dp[i][j] = s.charAt(j)...
private void getLPS(String s, int[] lps){ // i是后缀末尾的指针,j是前缀末尾的指针 int j = 0, i = 1; lps[0] = 0; // 从j=0,i=1开始找,错开了一位 while(i < s.length()){ // 如果字母相等,则继续匹配,最长相同前后缀的长度也加1 ...
int main(){ int i,j,n; while(scanf("%d",&n)!=EOF) { scanf("%s",a); for(i=n-1,j=0;i>=0;i--) b[j++]=a[i]; memset(p,0,sizeof(p)); for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(a[i-1]==b[j-1]) ...
public static boolean isPalindrome(String str) { int len = str.length(); for(int i=0; i<len/2; i++) { if(str.charAt(i)!=str.charAt(len-i-1) { return false; } return true; } Run Code Online (Sandbox Code Playgroud) 但找到回文子串的有效方法是什么? java algorithm substring ...