public class Solution { public boolean isPalindrome(int x) { int palindrome=0; int revers=x; if(revers<0) return false; else{ while(revers>0){ int m=revers%10; palindrome=m+palindrome*10; revers=revers/10; } if(palindrome==x) return true; else return false; } } }...
Determine whether an integer is a palindrome. Do this without extra space. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra space. You could also try reversing an integer. However, if ...
十进制 classSolution{public:boolisPalindrome2(intx){//二进制intnum=1,len=1,t=x>>1;while(t){num<<=1;t>>=1;len++;}len/=2;while(len--){if((num&x==0)&&(x&1)!=0){return0;}x&=(~num);x>>=1;num>>=2;}return1;}boolisPalindrome(intx){//十进制if(x<0)return0;intnum...
class Solution { public: int a[100005]; bool isPalindrome(int x) { if(x<0) return false; int pos=0; while(x) { a[pos++]=x%10; x/=10; } for(int i=0,j=pos-1;i<j;i++,j--) { if(a[i]!=a[j]) { return false; } } return true; } }; 1. 2. 3. 4. 5. 6....
Leetcode 321. Create Maximum Number sdncomversion博客 文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 Tyan 2021/02/04 2980 LeetCode 0336 - Palindrome Pairs concatenationdistinctlistpalindromeunique Given a list of unique words, find all pairs of distinct indices (i, j) in the given list, so...
public class Solution { public String longestPalindrome(String s) { // Write your code here int len = s.length(); boolean[][] dp = new boolean[len][len]; int start = 0, end = 0; for (int i = 0; i < len; i++) {
(int i = p - 1; i >= 0; i--) { cnt[fail[i]] += cnt[i]; } } }; class Solution { public: vector<vector<int>> num; bool fun(int x, int y) { if(x < 0 && y < 3) { return false; } if (y == 3) { if (x == -1) return true; return false; } for (int...
链接:https://leetcode.com/problems/palindrome-number/#/description 难度:Easy 题目:9.Palindrome Number Determine whether an integer is a palindrome. Do this without extra space. Some hints: Could negative integers be palindromes? (ie, -1) ...
输入:s = "leetcode" 输出:5 解释:插入 5 个字符后字符串变为 "leetcodocteel" 。 提示: 1 <= s.length <= 500 s 中所有字符都是小写字母。 思路: 动态规划 - 最长子序列 将字符串 s 反序得到 s' 设s 与 s‘ 的最长子序列为 m
9. 回文数 - 给你一个整数 x ,如果 x 是一个回文整数,返回 true ;否则,返回 false 。 回文数是指正序(从左向右)和倒序(从右向左)读都是一样的整数。 * 例如,121 是回文,而 123 不是。 示例 1: 输入:x = 121 输出:true 示例 2: 输入:x = -121 输