LeetCode 刷题记录 9. Palindrome Number
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
Input: 121
Output: true
Example 2:
Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:
Coud you solve it without converting the integer to a string?
方法1：
很容易想到反转数字 然后进行比较 但是会有溢出的问题，所以我们使用long long 存储反转后的数字
c++：
class Solution {
public:
bool isPalindrome(int x) {
int temp = x;
if(x < 0) return false;
//if(x == 0) return true;
long long reverse = 0;
while(temp){
reverse = reverse * 10 + (temp % 10);
temp /= 10;
}
bool flag;
flag = (reverse == x ) ? true : false;
return flag;
}
};
方法2：
不断取第一位和最后一位比较 如果相等 则继续 直到比较完成或者不一致为止
首先要找到最大的除数div 比如1991，div为1000
然后找到最左边一位 int left = x / div;
最右边一位 int right = x % 10; 判断是否相等
如果不相等直接返回 相等继续比较
x = (x % div) / 10;起到插头去位的作用 由于少了两位 div 要除以100
class Solution {
public:
bool isPalindrome(int x) {
if(x < 0) return false;
int div = 1;
while(x / div >= 10) div *= 10;
while(x > 0){
int left = x / div;
int right = x % 10;
if(left != right) return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
};
java：
class Solution {
public boolean isPalindrome(int x) {
if(x < 0) return false;
int div = 1;
while(x / div >= 10) div *= 10;
while(x > 0){
int left = x / div;
int right = x % 10;
if(left != right) return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
}
Python：
python中是False和True
class Solution(object):
def isPalindrome(self, x):
"""
:type x: int
:rtype: bool
"""
if x < 0: return False;
div = 1
while x / div >= 10: div *= 10;
while x > 0:
left = x / div
right = x % 10
if left != right: return False
x = (x % div) / 10
div /= 100
return True
方法3：
首先特例：x的尾数为0且本身不为0 肯定不是回文数予以排除
然后将数字的后半部分反转与前半部分比较
分两种情况x长度是奇数 例如19191 此时reverse为191 x为19 故判断x == reverse / 10
x长度是偶数 例如1991 此时reverse为19 x为19 故判断x == reverse
C++：
class Solution {
public:
bool isPalindrome(int x) {
if(x < 0 || (x != 0 && x % 10 == 0)) return false;
int reverse = 0;
while(x > reverse){
reverse = reverse * 10 + x % 10;
x /= 10;
}
return (x == reverse) || (x == reverse / 10);
}
};
java：
class Solution {
public boolean isPalindrome(int x) {
if(x < 0 || (x != 0 && x % 10 == 0)) return false;
int reverse = 0;
while(x > reverse){
reverse = reverse * 10 + x % 10;
x /= 10;
}
return (x == reverse) || (x == reverse / 10);
}
}
Python：
class Solution(object):
def isPalindrome(self, x):
"""
:type x: int
:rtype: bool
"""
if x < 0 or (x != 0 and x % 10 == 0): return False
reverse = 0
while x > reverse:
reverse = reverse * 10 + x % 10
x /= 10
return (x == reverse) or (x == reverse / 10)
方法4：
换一种想法 如果是回文数 他的回文数还是他自己 故不会溢出 溢出即不是回文数
即可调用之前已经写好的函数
c++：
class Solution {
public:
bool isPalindrome(int x) {
if(x < 0 || (x != 0 && x % 10 == 0)) return false;
return x == reverse(x);
}
int reverse(int x) {
int ans = 0;
//cout << (INT_MAX % 10) << (INT_MIN % 10);
while(x != 0){
int pop = x % 10;
if(ans > INT_MAX / 10 || (ans == INT_MAX / 10 && pop > (INT_MAX % 10))) return 0;
if(ans < INT_MIN / 10 || (ans == INT_MIN / 10 && pop < (INT_MIN % 10))) return 0;
ans = ans * 10 + pop;
x /= 10;
}
return ans;
}
};
java：
class Solution {
public boolean isPalindrome(int x) {
if(x < 0 || (x != 0 && x % 10 == 0)) return false;
return x == reverse(x);
}
public int reverse(int x) {
int ans = 0;
while(x != 0){
int remainder = x % 10;
int newAns = ans * 10 + remainder;
if((newAns - remainder) / 10 != ans) return 0;
ans = newAns;
x /= 10;
}
return ans;
}
}
python:
class Solution(object):
def isPalindrome(self, x):
"""
:type x: int
:rtype: bool
"""
if x < 0 or (x != 0 and x % 10 == 0): return False
return x == self.reverse(x)
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
factor = -1 if x < 0 else 1
if factor == 1:
reverseX = int(str(x)[::-1])
else:
reverseX = int(str(x)[1:][::-1])
return 0 if reverseX > (pow(2, 31) - 1) else factor * reverseX