[LeetCode]Longest Palindromic Substring

class Solution {//insert special character, then enumerate every character in workStr and record the max substring O(n^2)public:string longestPalindrome(string s) {// Start typing your C/C++ solution below// DO NOT write int main() functionstring workStr;workStr.resize(s.size()*2+1);for (int i = 0, j = 0; i < workStr.size(); ++i){if(i%2 == 0)workStr[i] = '_';else workStr[i] = s[j++];}//find the maximum palindromicint maxLen = 0;int maxLeftIdx;int maxRightIdx;for (int i = 0; i < workStr.size(); ++i){int leftIdx = i;int rightIdx = i;while (leftIdx > 0 && rightIdx < workStr.size()-1){if (workStr[leftIdx-1] == workStr[rightIdx+1]){leftIdx--;rightIdx++;}else break;}if (rightIdx-leftIdx > maxLen){maxLen = rightIdx-leftIdx;maxLeftIdx = leftIdx;maxRightIdx = rightIdx;}}maxLeftIdx /= 2;maxRightIdx /= 2;return s.substr(maxLeftIdx, maxRightIdx-maxLeftIdx);}};

second time

class Solution {//O(n), using the symmetric property of palindromepublic:    string longestPalindrome(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<char> workStr(2*s.size()+1);        for(int i = 0, k = 0; i < workStr.size(); ++i)//...        {            if(i%2 == 0) workStr[i] = '\0';            else workStr[i] = s[k++];        }                //populate p        vector<int> p(workStr.size(), 0);        int id = 0, mx = 0;        for(int i = 1; i < workStr.size(); ++i)        {            p[i] = mx > i ? min(p[id*2-i], mx-i) : 1;            while(i-p[i] >= 0 && i+p[i] < workStr.size()                   && workStr[i+p[i]] == workStr[i-p[i]]) p[i]++;            if(i+p[i]-1 > mx)            {                mx = i+p[i]-1;                id = i;            }        }        //select the max         int maxLen = 0;        int maxIdx = 0;        for(int i = 0; i < p.size(); ++i)            if(maxLen < p[i]-1) maxLen = p[i]-1, maxIdx = i;                int startIdx = (maxIdx-maxLen)/2;        return s.substr(startIdx, maxLen);    }};