題目：

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.

鏈接：?http://leetcode.com/problems/palindrome-number/

題解：

可以用除法和模運(yùn)算rebuild一個結(jié)果，然后比較和輸入是否相等。這里其實(shí)有可能會overflow，就是在 result * 10的時候，但越界的話也會返回一個32位整數(shù)，這個整數(shù)是 result * 10結(jié)果轉(zhuǎn)換為64位里面的low 32位，一般來說與輸入不同。所以結(jié)果返回false，也能過AC，但是不嚴(yán)謹(jǐn)。

Time Complexity - O(logx)， Space Complexity - O(1)。

public class Solution {public boolean isPalindrome(int x) {if(x == 0)return true;int result = 0, temp = x; while(temp > 0){result = 10 * result + temp % 10;temp /= 10;}return result == x;} }

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另外一種方法可以避免overflow。先用對數(shù)計算出x有幾位，然后通過數(shù)學(xué)運(yùn)算比較第一位和最后一位是否相等，接下來去掉最大的一位和最小的一位，再將digitNum - 2，繼續(xù)進(jìn)行計算。

Time Complexity - O(logx)， Space Complexity - O(1)。

public class Solution {public boolean isPalindrome(int x) {if(x < 0)return false;int digitNum= (int)(Math.log(x) / Math.log(10)); int left = 0, right = 0;while(x > 0){int powInTens = (int)Math.pow(10, digitNum);left = x / powInTens;right = x % 10;if(left != right)return false;x -= left * powInTens;x /= 10;digitNum -= 2 ;}return true;} }

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二刷：

Java: - Reverse all digits:

public class Solution {public boolean isPalindrome(int x) {if (x < 0) {return false;}int res = 0, tmp = x;while (tmp > 0) {res = res * 10 + tmp % 10;tmp /= 10;}return res == x;} }

Reverse half digits:

public class Solution {public boolean isPalindrome(int x) {if (x < 0 || (x != 0 && x % 10 == 0)) {return false;}int res = 0;while (x > res) {res = res * 10 + x % 10;x /= 10;}return (res == x) || (x == res / 10);} }

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Python:

class Solution(object):def isPalindrome(self, x):""":type x: int:rtype: bool"""if x < 0 or (x != 0 and x % 10 == 0):return Falseres = 0while x > res:res = res * 10 + x % 10x /= 10return (x == res) or (x == res / 10)

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三刷:

Java:

計算全部digits: ?設(shè)立一個int n = x， 然后我們計算一下n的按位反轉(zhuǎn)數(shù)res，最后比較 n == res

public class Solution {public boolean isPalindrome(int x) {if(x < 0) {return false;}int n = x;int res = 0;while (x > 0) {res = res * 10 + x % 10;x /= 10;}return res == n;} }

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計算一半digits:

這里我們要先剪掉特殊情況 x % 10 == 0，這時候假如x不為0的話，肯定不是palindromic number。

之后while循環(huán)作比較的條件是 ?x ?> res。 我們只需要計算一半的digits，然后比較是否 x == res， 或者 x == res / 10。

當(dāng)x是偶數(shù)長度的話，我們比較 x == res; 當(dāng)x時奇數(shù)長度的時候，比如 x = 1，這時我們比較的是 x == res / 10。合起來用一個或運(yùn)算就可以了。

public class Solution {public boolean isPalindrome(int x) {if(x < 0 || (x != 0 && x % 10 == 0)) {return false;}int res = 0;while (x > res) {res = res * 10 + x % 10;x /= 10;}return (x == res || x == res / 10) ;} }

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Reference:

https://leetcode.com/discuss/33500/an-easy-lines-code-only-reversing-till-half-and-then-compare

https://leetcode.com/discuss/23563/line-accepted-java-code-without-the-need-handling-overflow

https://leetcode.com/discuss/12693/neat-ac-java-code-o-n-time-complexity?

https://leetcode.com/discuss/65915/python-solution-with-other-variable-introduced-besides-input

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轉(zhuǎn)載于:https://www.cnblogs.com/yrbbest/p/4430410.html

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