it: Public class solution{ PublicBOOL Ispalindrome (intx) {if(X 0)return false;//Determine the length of x, for example x=232,div is equal to int Div=1; while(X/Div>=Ten)Div*=Ten; while(X! =0) {//Left start Count intleft = x/Div;//Right Start Count intright = x%Ten;if(Left! = right)return false; x = (x%Div) /Ten;Div/= -; }return true; }}Performance is not satisfactory, but the same code in Java, it seems that C # faster.1150611506 test cases passed.Statu
Give a number, well, it's not a number, it's a bunch of numbers,
Like 98237482340328490328490324893024, a very long number.
To find the next palindrome, this palindrome is larger than the current value (not equal to).
Such as:
9 9 9 9->1 0 0 0 1
1 2 3 4 5->1 2 4 2 1
The algorithm time efficiency is O (n), where n refers to the number of digits, not the number, such as 123456789, only 9 bits, then the
Title: http://www.lydsy.com/JudgeOnline/problem.php?id=2565Record the longest extension of each point forward, both positive and negative, and enumerate the split points.#include #include#include#include#defineRep (i,l,r) for (int i=l;i#defineDown (i,l,r) for (int i=l;i>=r;i--)#defineCLR (x, y) memset (x,y,sizeof (×))#definell Long Long#defineMAXN 100500using namespacestd;intlen[maxn],cnt[maxn],p1[maxn],p2[maxn],fail[maxn],s[maxn],to[maxn][ -];intN,ans,tot,last,l,len1,len2;CharCH1[MAXN],CH2[MAXN
Gram. Inputthe input consists of a single line, which contains a string of Latin alphabet letters (no other characters would appe AR in the string). String length would not exceed characters. Outputthe longest substring with the mentioned property. If There is several such strings you should output the first of them. Sample
input
Output
Thesampletextthatcouldbereadedthesameinbothordersarozaupalanalapuazora
Arozaupalanalapuazora
The code is as
public class stringproblem{//manacher algorithm preprocessing public static char[] Manacherstring (String str) {char[
] Chararr = Str.tochararray ();
char[] res = new Char[str.length () * 2 + 1];
int index = 0; for (int i = 0; I!= res.length i++) {Res[i] = (I 1) = = 0?
' # ': chararr[index++];
return res; //manacher algorithm O (N) complexity to find the longest palindrome substring public static int maxlcpslength
Problem G: Concave digitsDescriptionDefine a number called a concave number, that is, from high to position, the number of each digit first does not increment again and again is not descending, and the number is a palindrome number, that is, from left to right and read from right to left is the same, only to form a concave peak, such as 543212345,5544334455 is a valid concave number , 543212346,123321,111111 is not a concave number. Now ask you [l,r]
, for palindrome number, the definition of Baidu is as follows:"Palindrome" is to read the anti-reading can read through the sentence, it is all the time and time there is a rhetorical way and word game, such as "I for everyone, everyone for me" and so on.In mathematics there is such a characteristic of such a number that it becomes a
the interval. Now the interval (i, j) used by the sub-range, before all have been finished, so or that sentence, direct use is good.The opinion of that place is different to the welcome point, learn from each other!!#include #include #include String.h>#include UsingNamespaceStdConstint MOD =100007;int T, Len, d[1010][1010];Char s[1010];voidDP () {Forint i =1; i ) {ForInt J =0; J ) {if (s[j] = = s[j+I]) d[j][i+j] = (d[j][i+j-1] + d[j+1][I+J] +1)%MOD;E
Determine whether an integer is a palindrome. Do this without extra space. (source)Determines whether a number is a palindrome number, and does not use additional storage space."Palindrome" is to read the anti-reading can read through the sentence, then palindrome number, it
I. Description of the problem
Palindrome string and palindrome sequence are different:
SUBSTRING, be sure to continueSub sequence, not necessarily continuous
In fact, the longest palindrome string can be converted into LCS to do, the specific method is:
Generates a reverse string of the original string and then uses a DP to find the LCS for the original string a
] represents the longest palindrome length at the end or beginning of the current point ...There are two ways to find LR ... One is my relatively low method of seeking ... One is lty big guy side do manacher side to seek LR%%% procedure ...I will not say, look at the code ... Expressive Power 2333 ...Lty said can be in manacher violent match time (that is the following sentence ...)while (Str[i-p[i]]==str[i
(1) Using recursive method to determine whether a string is a palindrome ( palindrome);" A palindrome is the same sentence that is read and read in reverse. Like , "Who am I?"The algorithm for detecting palindrome using recursive algorithm is described as follows:A sing
"Palindrome" is a sentence to read and read through. It is a rhetorical way and word games, such as "I for everyone, everyone for me" and so on. In mathematics there are also such a number of numbers that have this characteristic and become palindrome numbers (palindrome number).set N is a random natural number. If the
DescribeSmall hi and small ho is a pair of good friends, born in the information society, they have a great interest in programming, they agreed to help each other, in the programming of learning along the road together.On this day, they encountered a string of strings, so little hi to small ho raised the classic question: "Little ho, you can find each of them in these strings each of the longest palindrome string it?" " Little Ho asked curiously, "W
1 Palindrome PartitioningProblem Source: Palindrome PartitioningThe problem is simply given a string, dividing the string into multiple parts, satisfying each part as a palindrome, and outputting all possible scenarios.The difficulty of the problem is relatively large, it is likely that the first encounter without thinking, it is normal. Below we analyze a little
Turn from blog: www.cnblogs.com/mickole/articles/3578298.html
Title: (Alternative topics can be found in the Pat ladder exercises, of course, the problem N3 can be too)
A string of length n (n Large), which is the longest palindrome in the string. (Baidu 2014 School Recruit written questions) The topic pointed out "N is very big", that is, we do not want to go through the way to find this string, I think of a solution, time complexity should not be h
Manacher algorithm (Http://www.jianshu.com/p/799bc53d4e3d)For a long string, the time complexity of O (n^2) is unacceptable. Can We do better?Let's take a look at the flaw in Solution 2 .1) due to the odd-even nature of palindrome string length, the symmetry axis position of different properties is caused.Solution 2To handle the two cases separately, 2) Many substrings are repeatedly accessed, resulting in poor time efficiency.Defect 2) can be through
" Questions Eyes " Leetcode (9 ) : palindrome numberurl:https://leetcode.com/problems/palindrome-number/DescriptionDetermine whether an integer is a palindrome. Do this without extra space."Chinese description"Determines whether an integer is a palindrome number, similar to 121, 2332, 1000001, 22. doing this does not a
This article is introduced to Manacher in detail, and it is easy to read, the original is transferred from: http://blog.csdn.net/pi9nc/article/details/9251455First, the problem descriptionGiven a known string str[], now want to find a longest palindrome substring (positive and backward sequential reads) within the time complexity of O (n).Manacher first found that the time complexity of O (n) can be used to solve this problem, so this method is called
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