【leetcode】Longest Palindromic Substring

From : https://leetcode.com/problems/longest-palindromic-substring/

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

Solution 1 : 

动态规划。 isPalin 表示i到j之间是否是回文，时间O(n^2)， 空间O(n^2)。

class Solution {public:    string longestPalindrome(string s) {        if(s == "") return "";        int len=s.size(), start=0, end=0;        bool isPalin[len][len];        // init        for(int i=0; i<len; i++) {            for(int j=0; j<=i; j++) {                isPalin[i][j] = true;            }            for(int j=i+1; j<len; j++) {                isPalin[i][j] = false;            }        }                // core        for(int j=1; j<len; j++) {            for(int i=0; i<j; i++) {                if(s[i] == s[j]) {                    isPalin[i][j] = isPalin[i+1][j-1];                    if(isPalin[i][j] && j-i > end-start) {                        start = i;                        end = j;                    }                }            }        }                return s.substr(start, end-start+1);    }};

Solution 2:

回文有中心，遍历这些中心。时间O(n^2)，空间O(1)。

class Solution {public:    void expendFromCenter(string &str, int i) {        int st, ed;        ed = (i>>1)+1;        st = ed - 1 - (!(i&1));        while(st >= 0 && ed < length && str[st] == str[ed]) {            st--; ed++;        }st++; ed--;        if(ed - st > end - start) {            start = st;            end = ed;        }    }    string longestPalindrome(string s) {        if((length = s.size()) <= 1) return s;start = end = 0;        for(int i=0, n=2*length-1; i<n; i++) {            expendFromCenter(s, i);        }        return s.substr(start, end-start+1);    }private:    int start, end, length;};

剪枝处理，求取可扩展的半径r，只对可能的情况做处理，半径超出字符串界或者半径边界本身不相等，那么不可能成为更长的回文，剪枝。

class Solution {public:    int find(string& s, int c, int& st, int& ed, int& len) {        int i, j;        if(c&1) {            i = c>>1;            j = i+1;        } else {            i = (c>>1)-1;            j = i+2;        }        int r = (ed-st-j+i+1)>>1;        if(i-r < 0 || j+r >= len || s[i-r] != s[j+r]) {            return r;        }        while(i>=0 && j<len && s[i] == s[j]) {            if(j-i > ed-st) {                st = i;                ed = j;            }            --i;            ++j;        }        return r;    }    string longestPalindrome(string s) {        if(s.size() <= 1) return s;        int start = 0, end = 0, len = s.size();        int centers = len+len-1;        for(int i=1, r=0; i<centers-r; ++i) {            r = find(s, i, start, end, len);        }        return s.substr(start, end-start+1);    }};

Solution 3：

Manacher’s Algorithm算法，详见

http://articles.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html 

http://www.felix021.com/blog/read.php?2040

根据回文的对称性，选出最大回文。

class Solution {public:// Transform S into T.// For example, S = "abba", T = "^#a#b#b#a#$".// ^ and $ signs are sentinels appended to each end to avoid bounds checkingstring preProcess(string &s) {int n = s.length();if (n == 0) return "^$";string ret = "^";for (int i = 0; i < n; i++)ret += "#" + s.substr(i, 1); ret += "#$";return ret;} string longestPalindrome(string s) {string T = preProcess(s);int n = T.length();int *P = new int[n];int C = 0, R = 0;for (int i = 1; i < n-1; i++) {int i_mirror = 2*C-i; // equals to i' = C - (i-C)    P[i] = (R > i) ? min(R-i, P[i_mirror]) : 0;    // Attempt to expand palindrome centered at iwhile (T[i + 1 + P[i]] == T[i - 1 - P[i]])P[i]++; // If palindrome centered at i expand past R,// adjust center based on expanded palindrome.if (i + P[i] > R) {C = i;R = i + P[i];}} // Find the maximum element in P.int maxLen = 0;int centerIndex = 0;for (int i = 1; i < n-1; i++) {if (P[i] > maxLen) {maxLen = P[i];centerIndex = i;}}delete[] P;  return s.substr((centerIndex - 1 - maxLen)/2, maxLen);}};