Title: Longest palindrome string (c + +)

See the first reaction to this question is done ah, and then began to write, and so finish a test. Emmmmm, originally is the longest palindrome string is not the longest palindrome subsequence, but write is written, I will change the code slightly to adapt to the topic, the code is as follows:

Static ConstAuto IO_SPEED_UP =[] () {Std::ios::sync_with_stdio (false);    Cin.tie (nullptr); return 0;} ();classSolution { Public:    stringLongestpalindrome (strings) {intn =s.length (); if(n = =1|| n = =0)            returns; intMax =0; intBegin =0; Vector<vector<int> > DP (n,vector<int>(n));  for(intj =0; J < N; J + +) {Dp[j][j]=1;  for(inti = J-1; I >=0; i--)            {                if(S[i]! =S[j])Continue; if(Dp[i +1][j-1] ==0&& I! = J-1)                    Continue; DP[I][J]= Dp[i +1][j-1] +2; if(Max >=Dp[i][j])Continue; Max=Dp[i][j]; Begin=i; }        }        if(max = =0)            returnS.SUBSTR (Begin,1); Else            returns.substr (begin, Max); }};

Then this code a look is very rubbing ah, completely ill should question, guessing efficiency must be very low, and so on after the submission, sure enough, only more than 13.74% of the code. Had to delete the rewrite.

The second write code is more efficient than the 32.9% code below:

Static ConstAuto IO_SPEED_UP =[] () {Std::ios::sync_with_stdio (false);    Cin.tie (nullptr); return 0;} ();classSolution { Public:    stringLongestpalindrome (strings) {intn =s.length (); if(n = =1|| n = =0)            returns; intMax =0; intdiff =0; intBegin =0; intEnd =0; intMaxbegin =0; intMaxend =0;  for(inti =0; I < n; ++i) { for(intj = N-1; J >0; --j) {if(Max > J-i) Break; Begin=i; End=J;  while(S[begin] = = S[end] && begin <=end) {                    ++begin; --end; }                if(End < begin && Begin-end <3) {diff= J-i; if(Max <diff) {Max=diff; Maxbegin=i; Maxend=J; }                }            }            if(Max > N-i-1)                 Break; }        returnS.substr (Maxbegin, Max +1); }};

Well, I feel like this is the best solution I can figure out, and then I looked at the solution on the website. which

The second-to-last solution is this one.

classSolution { Public:    stringLongestpalindrome (strings) {intLen =0; intleft =0; intright =0; stringresult =""; inti =0;  while(I <s.size ()) { Left=i; Right=i;  while(Right +1< S.size () && s[right] = = S[right +1]) right++; I= right +1;  while(Left >=0&& Right < S.size () && s[left] = =S[right]) { Left--; Right++; }            if(Len < Right-left-1) {len= Right-left-1; Result= S.substr (left +1, Len); }        }        returnresult; }};

You didn't. Optimize the input and output stream, and then split the efficiency of the result string multiple times, it can achieve 6ms execution Speed row second.

I looked at the code of the Great God and found that the main idea was to start with a central subscript, and then he used a clever way to deal with even-length strings starting with two characters, that is, after the center is selected, if the character after the loop is typeface equal to the selected word, then +1. This way, if an even-length string is present, left and right point to the adjacent two characters at the beginning. I added this big god code to the speed, and then dropped to the leetcode on the detection, the modified code is as follows:

Static ConstAuto IO_SPEED_UP =[] () {Std::ios::sync_with_stdio (false);    Cin.tie (nullptr); return 0;} ();classSolution { Public:    stringLongestpalindrome (strings) {intLen =0; intleft =0; intright =0; intMaxres =0; intLeftres =0; inti =0;  while(I <s.size ()) { Left=i; Right=i;  while(Right +1< S.size () && s[right] = = S[right +1]) right++; I= right +1;  while(Left >=0&& Right < S.size () && s[left] = =S[right]) { Left--; Right++; }        if(Len < Right-left-1) {len= Right-left-1; Leftres=Left ; Maxres=Len; }    }    returnS.SUBSTR (Leftres +1, Maxres); }};

The test results are then displayed as follows:

More than 100% and show execution speed of 0ms, bad.

After a general look at the first row of the algorithm, found that the "#" to fill the character as the original 2n+1 and then from the central character calculus, completely without the brother's ingenious, here will not mention.

Title: Longest palindrome string (c + +)