Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using ext…

Determine whether an integer is a palindrome. Do this without extra space. 这题貌似解法挺多,直接用简单的把数倒置,没有考虑数据溢出的问题,如需解决可以把转置数设为long long即可.另外方法可以用到前后匹配.递归比较等. class Solution { public: bool isPalindrome(int x) { int a=x; ; ) return false; while(x) { n*=; n+=x…

题目: 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…

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…

Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using ext…

class Solution(object): def isPalindrome(self, x): """ :type x: int :rtype: bool """ if x < 0: return False x=str(x) i=0 j=len(x)-1 while i < j: if x[i] != x[j]: return False i+=1 j-=1 return True…

一次AC 题目要求中有空间限制,因此没有采用字符串由量变向中间逐个对比的方法,而是采用计算翻转之后的数字与x是否相等的方法: class Solution: # @return a boolean def isPalindrome(self, x): o = x ret = 0 flag = 1 if x < 0: return False while(x!=0): ret = ret*10+x%10 x = x/10 return ret == o…

class Solution { public: bool isPalindrome2(int x) {//二进制 int num=1,len=1,t=x>>1; while(t){ num<<=1; t>>=1; len++; } len/=2; while(len--){ if((num&x==0)&&(x&1)!=0){ return 0; } x&=(~num); x>>=1; num>>=2; }…

本来推断回文串是一件非常easy的事情,仅仅须要反转字符串后在与原字符串相比較就可以.这道题目明白说明不能使用额外的空间.那么使用将其分解连接成字符串的方法便不是可行的.仅仅好採用数学的方式: 每次取最高位和最低位相比較,总的位数能够用一个while先处理出来,循环直至取余和除数相等. 详细见代码: class Solution { public: bool isPalindrome(int x) { if(x<0) //special due return false; if(x<10) r…

目录 题目链接 注意点 解法 小结 题目链接 Palindrome Number - LeetCode 注意点 负数肯定是要return false的 数字的位数要分奇数和偶数两种情况 解法 解法一:将数字转化为字符串,然后对字符串从中间开始往两边拓展比较,要分长度为奇数和偶数的情况.时间复杂度为O(n) class Solution { public: bool isPalindrome(int x) { if(x<0 || (x%10 == 0 && x!=0)) { retur…