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 you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
There is a more generic way of solving this problem.
Normal Way
reverse the Integer with long Integer
1 2 3 4 5 6 7 8 9 10 11 12 13 14 | class Solution {public: bool isPalindrome(int x) { if(x < 0) return false; long num = x,result = 0; while(num){ result = result * 10 + num %10; num/=10; } if( result == x) return true; else return false; }}; |
Improved Way
What if the input number is already a long integer?
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | class Solution {public: bool isPalindrome(int x) { if(x<0) return false; if(x<10) return true; if(x%10==0) return false; //actual logic int v=x%10; x=x/10; while(x-v>0)//x is getting smaller every step, { v=v*10+x%10; x/=10; } if(v>x){v/=10;} return v==x?true:false; }}; |
比如
奇数长度：1234321
每次通过除以10和对10取余来保存下前半部分x和后半部分v，比如 x = 123432, v=1;
直到x=123，v=1234的时候，循环结束，再通过 v/10 == x，来判断数是不是回环的。
偶数长度：则直接比较 v==x
Genetic Way
compare the first and last number of the input x.
Maximum comparing time will be half of the length of x
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | class Solution {public: bool isPalindrome(int x) { if (x < 0) return false; int d = 1; // divisor while (x / d >= 10) d *= 10; // 计算出是多少位数 while (x > 0) { int q = x / d; // quotient int r = x % 10; // remainder if (q != r) return false; x = x % d / 10; d /= 100; } return true; }}; |