https://leetcode-cn.com/problems/palindrome-number/
Palindrome Number
Determine whether an integer is a palindrome number. The number of palindromes refers to integers that read in the same order (from left to right) and reverse order (from right to left).
Example 1:
Input: 121
Output: true
Example 2:
Input: -121
Output: false
Explanation: Reading from left to right is -121. From right to left, it is 121-. Therefore it is not a palindrome.
Example 3:
Input: 10
Output: false
Explanation: From right to left, it is 01. Therefore it is not a palindrome.
Solution 1
8ms 6.1mb
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0) {
return false;
}
long int res = 0, my_x = x;
while (x != 0) {
int d = x % 10;
res = res * 10 + d;
x = x / 10;
}
if (res == my_x) {
return true;
} else {
return false;
}
}
};
Solution 1.5
A judgment [0, 10)is added at 20ms 5.9mb , it can directly return true, if there is something similar to 100, 1000, etc. directly returns false.
Compared to solution one, I initially thought it would be faster, so I didn't need to loop.
In fact, it is also time-consuming to judge the need to calculate in the test case.
class Solution {
public:
bool isPalindrome(int x) {
if (x < 10 && x >= 0) {
return true;
}
if (x < 0 || x % 10 == 0) {
return false;
}
long int res = 0, my_x = x;
while (x != 0) {
int d = x % 10;
res = res * 10 + d;
x = x / 10;
}
printf("res %d\n", res);
return res == my_x;
}
};
Solution 2
Convert to a string to solve.
Of course, you can also use the stack, but I don't need it, just judge.
If you use the stack, you have to go through O (n). I only need to go through O (n / 2) in this way, but the time complexity is actually O (n).
28ms, 6mb
class Solution {
public:
bool isPalindrome(int x) {
if (x < 10 && x >= 0) {
return true;
}
if (x < 0 || x % 10 == 0) {
return false;
}
std::string x_str = std::to_string(x);
int start = 0;
int end = x_str.size() - 1;
while (start < end) {
if (x_str[start] != x_str[end]) {
return false;
}
start ++;
end --;
}
return true;
}
};
EOF
