Palindrome Number
2015-08-01 17:15
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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 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.
题目比较简单,就直接贴代码,还是注意溢出与反转的技巧。
class Solution {
public:
//这跟反转整数一样的,反转之后直接比较是否一样,注意溢出的判断
bool isPalindrome(int x) {
if(x < 0|| x >= INT_MAX)
return false;
long long tmpSum = 0;
int tmpbit = 0;
int tmpx = x;
while(tmpx)
{
tmpbit = tmpx % 10;
tmpx = tmpx / 10;
tmpSum *= 10;
tmpSum += tmpbit;
}
if(tmpSum >= INT_MAX || tmpSum != x)
return false;
return true;
}
};
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 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.
题目比较简单,就直接贴代码,还是注意溢出与反转的技巧。
class Solution {
public:
//这跟反转整数一样的,反转之后直接比较是否一样,注意溢出的判断
bool isPalindrome(int x) {
if(x < 0|| x >= INT_MAX)
return false;
long long tmpSum = 0;
int tmpbit = 0;
int tmpx = x;
while(tmpx)
{
tmpbit = tmpx % 10;
tmpx = tmpx / 10;
tmpSum *= 10;
tmpSum += tmpbit;
}
if(tmpSum >= INT_MAX || tmpSum != x)
return false;
return true;
}
};
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