LeetCode 5 Longest Palindromic Substring (maximum echo Substring)

LeetCode 5 Longest Palindromic Substring (maximum echo Substring)

Translation

Given a string S, find its maximum echo substring. You can assume that the maximum length of S is 1000, and there is a unique maximum echo substring.

Original

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.

Brute force search, O (n3)

public static string LongestPalindrome(string s){    int len = s.Length;    for (int i = len; i > 0; i--)        for (int j = 1; j <= len + 1 - i; j++)            if (isPalindrome(s.Substring(j - 1, i)))                return s.Substring(j - 1, i);    return null;}public static bool isPalindrome(string s){    for (int i = 0; i < s.Length / 2; i++)        if (s.Substring(i, 1) != s.Substring(s.Length - 1 - i, 1))            return false;    return true;}

Dynamic Planning, time: O (n2) , Space: O (n2)

Public class Solution {public string LongestPalindrome (string s) {int sLen = s. length; int lonBeg = 0; int maxLen = 1; bool [,] DP = new bool [1000,100 0]; for (int I = 0; I <sLen; I ++) {DP [I, I] = true;} for (int I = 0; I <sLen-1; I ++) {if (s [I] = s [I + 1]) {DP [I, I + 1] = true; lonBeg = I; maxLen = 2 ;}} for (int len = 3; len <= sLen; len ++) // All substrings whose lengths start from 3 {for (int I = 0; I <sLen + 1-len; I ++) {int j = len-1 + I; // j is the index at the end of the array. if (s [I] = s [j] & DP [I + 1, j-1]) {DP [I, j] = true; // from I to j, return lonBeg = I; // lonBeg is the starting index, which is equal to I maxLen = len; // maxLen is the string length }}return s. substring (lonBeg, maxLen );}}

However, the tragedy continues ......

Submission Result: Memory Limit Exceeded

public class Solution{    public string LongestPalindrome(string s)    {        int nLen = s.Length;        if (nLen == 0) return ;        string lonStr = s.Substring(0, 1);        for (int i = 0; i < nLen - 1; i++)        {            string p1 = ExpandAroundCenter(s, i, i);            if (p1.Length > lonStr.Length)                lonStr = p1;            string p2 = ExpandAroundCenter(s, i, i + 1);            if (p2.Length > lonStr.Length)                lonStr = p2;        }        return lonStr;    }    public static string ExpandAroundCenter(string s, int n, int m)    {        int l = n, r = m;        int nLen = s.Length;        while (l >= 0 && r <= nLen - 1 && s[l] == s[r])        {            l--;            r++;        }        return s.Substring(l + 1, r - l - 1);    }}

Okay, this method comes from the network ...... It is said that O (n2) Time, but only O (1) Space!