python 检查回文的偶数/奇数？
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Check even/odd for Palindrome?
提问by eozzy
Is it a good idea to check for odd/even length of a palindrome number/string? Most snippets I came across don't do this basic test. If length is even, it can't be a palindrome, no?
检查回文数/字符串的奇数/偶数长度是个好主意吗？我遇到的大多数片段都没有进行这个基本测试。如果长度是偶数，它就不可能是回文，不是吗？
if len(var) % 2 != 0:
# could be a palindrome, continue...
else:
break
Or is it just better (i.e faster) to start comparing the first and last numbers/letters directly?
或者直接开始比较第一个和最后一个数字/字母是否更好（即更快）？
Edit: Okay, stupid question, should've thought twice! :)
编辑：好吧，愚蠢的问题，应该三思而后行！:)
回答by TabbyCool
ABBA- an example of palindromeof four letters meaning even length.
A palindromeis a word, phrase, number, or other sequence of characterswhich reads the same backward or forward...
回答by sth
The easiest way to check for a palindrome is to simply compare the string against it's reverse:
检查回文的最简单方法是简单地将字符串与其反向进行比较：
def ispalindrome(s):
return s == s[::-1]
This uses extended slices with a negative step to walk backwards through sand get the reverse.
这使用带有负步的扩展切片向后走s并获得相反的结果。
回答by Aly
baab = palindrome and has length of 4 which is even
baab = 回文，长度为 4，是偶数
回答by PaulMcG
Try this:
试试这个：
is_palindrome = lambda s : all(s1==s2 for s1,s2 in zip(s[:len(s)/2],s[-1:-(len(s)+1)/2:-1]))
only checks the front half with the back half, and short-circuits as soon as a mismatch is found.
只检查前半部分和后半部分，一旦发现不匹配就短路。
回答by Andrew Noyes
Simple case: aa.
简单案例：aa。
More complicated case: aaaa.
更复杂的情况：aaaa。
And so on.
等等。
回答by Attila O.
回答by Mohamed Yahya
If string.length is even Then : All chars count should be even, so we can not have a char with odd count.
如果 string.length 是偶数那么：所有字符数都应该是偶数，所以我们不能有一个奇数的字符。
If string.length is odd Then: One char count must be odd, so not all chars' count should be even.
如果 string.length 是奇数 那么：一个字符的计数必须是奇数，因此并非所有字符的计数都应该是偶数。
--------------- I implemented the following JavaScript for the follow up roles :
--------------- 我为后续角色实现了以下 JavaScript：
function isStrPermutationOfPalindrome(_str) { // backward = forward
var isPermutationOfPalindrome = true;
var _strArr = [..._str];
var _strArrLength = _strArr.length;
var counterTable = getCharsTabularFrequencies(_str);
var countOdd = 0;
var countEven = 0;
for (let [ky, val] of counterTable) {
if (val % 2 == 0) {
countEven = countEven + 1;
} else {
countOdd = countOdd + 1;
}
}
if (_strArrLength % 2 == 0) {
//Even count of all characters,otherwise false.
//so can not have a character with odd count.
if (countOdd != 0) {
isPermutationOfPalindrome = false;
}
} else {
//Odd count of 1 character
//so not all chars with even count, only one char of odd count.
if (countOdd > 1 || countOdd == 0) { //no odd, or more than one odd [ only one odd should be to return true]
isPermutationOfPalindrome = false;
}
}
return isPermutationOfPalindrome;
}
function getCharsTabularFrequencies(str) {
str = str.toLowerCase();
var arr = Object.assign([], str);
var oMap = new Map();
var _charCount = 0;
for (let i = 0; i < arr.length; i++) {
if (arr[i] === ' ') {
continue;
}
_charCount = 0;
for (let j = 1; j < arr.length; j++) {
{
if (arr[i] === arr[j]) {
_charCount = _charCount + 1;
}
}
}
if (i == 0)
_charCount = _charCount + 1;
if (!oMap.has(arr[i]))
oMap.set(arr[i], _charCount)
}
return oMap;
}
let _str = 'tactcoapapa';
console.log("Is a string of '" + _str + "' is a permutation of a palindrome ? ANSWER => " + isStrPermutationOfPalindrome(_str));
回答by RAHIMAN
n=raw_input("Enter a string==>")
n=int(n)
start=0
term=n
while n>0:
result=n%10
start=start*10+result
n=n/10
print start
if term==start:
print "True"
else:
print "False"
