Shortest Palindrome

Shortest Palindrome

Total Accepted: 13465 Total Submissions: 74427 Difficulty: Hard

Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.

For example:

Given

"aacecaaa"

, return

"aaacecaaa"

.

Given

"abcd"

, return

"dcbabcd"

.

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(M) Longest Palindromic Substring (E) Implement strStr()

先用manacher求出以第一个字符开头的最长回文的长度，然后把剩下的字符逆序放在原始串的前面

class Solution {
public:
void preProcess(string &s,int sSize,string& copedStr){
for(int i=0;i<sSize;i++){
copedStr.push_back('*');
copedStr.push_back(s[i]);
}
copedStr.push_back('*');
}
int longestPalindromeLengthStartWithFirstChar(string s) {
int sSize = s.size();
if(sSize<=1){
return sSize;
}
string str;
preProcess(s,sSize,str);
int strSize = 2*sSize+1;
vector<int> parr(strSize,1);
int mid = 0;
int left =-1,right = 1;
int pl = -1, pr = 1;
int radius = 0;
for(;right<strSize;right++){
radius = 0;
if(right>=pr){
while((right-radius>=0 && right+radius<strSize) && str[right-radius]==str[right+radius]){
radius++;
}
parr[right] = radius;
}else{
left = mid - (right-mid);
if(left - parr[left] +1 < pl+1){
parr[right] = left - pl;
}else if(left - parr[left] +1 > pl+1){
parr[right] = parr[left];
}else{
radius = parr[left]-1;
while((right-radius>=0 && right+radius<strSize) && str[right-radius]==str[right+radius]){
radius++;
}
parr[right] = radius;
}
}
radius = parr[right] ;
if(right + radius >= pr){
pr = right + radius ;
pl = right - radius ;
mid = right;
}
}
int maxLen = 1;
for(int i=1;i<strSize;i++){
if(i-parr[i]+1<=1){
int tmpRadius = i%2==0 ? parr[i]-1:parr[i];
if(tmpRadius>maxLen ){
maxLen = parr[i]-1;
}
}
}
return maxLen;
}
string shortestPalindrome(string s) {
int j = longestPalindromeLengthStartWithFirstChar(s);
string tmpStr;
cout<<j<<endl;
for(int i=s.size()-1;i>=j;i--){
tmpStr.push_back(s[i]);
}
return tmpStr+s;
}
};