LeetCode 5_Longest Palindromic Substring

LeetCode 5_Longest Palindromic Substring 

题目描写叙述：

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. 也即：求字符串中最长回文子串。

回文是什么我就不多少了。能够百度下！

方法一：暴力法（O(n^3)）

两层循环扫描字符串的全部子串，之后推断选出的字符子串是否是回文，若是则看其长度！

代码例如以下：

class Solution {
public:
    string longestPalindrome(string s) 
    {
        // 暴力法O(n^3)
        int n = s.size();
		if (n == 0 || n == 1)
			return s;
		int maxLength = 1;
		int k1 = 0, k2 = 0;
		for (int i = 0; i < n; i++)
		{
			for (int j = i + 1; j < n; j++)
			{
				int k = 0;
				int sign = 0;
				while (k < (j - i + 1)/2 && s[i + k] == s[j - k])
					k++;
				if(k == (j - i + 1) / 2 )
				{
					sign = 1;
					if (j - i + 1 > maxLength)
					{
						maxLength = j - i + 1;
						k1 = i;
						k2 = j;
						if (maxLength == n - i)
						    return s.substr(k1,k2+1);
					}
				}
			}
		}
		return s.substr(k1,k2+1);
}

因为题目说能够假定字符串的长度不超过1000，所以建立一个table[1000][1000] 的bool表，初始化为false。如果某子串（如果 i 到 j ）为回文。令table[ i ][ j ]为true。之后推断的时候先查表和更新表。代码例如以下：

class Solution {
public:
    string longestPalindrome(string s) 
    {
    		int n = s.length();
		if(n == 0 || n == 1)
		    return s;
		int maxLength = 1;
		int palindromBegin = 0;
		bool table[1000][1000] = {false};
		for(int i = 0; i < n; i++)
			table[i][i] = true;
		for (int i = 0; i < n; i++)
			if(s[i] == s[i + 1])
			{
			    table[i][i + 1] = true;
				maxLength = 2;
				palindromBegin = i;
			}
		for (int len = 3; len <= n ; len++)
		{
			for (int i = 0; i < n - len + 1; i++)
			{
				int j = i + len - 1;
				if (s[i] == s[j] && table[i + 1][j - 1] == true)
				{
				    table[i][j] = true;
					maxLength = len;
					palindromBegin = i;
				}
			}
		}
		return s.substr(palindromBegin, maxLength);
}

事实上还能够考虑回文的中心点。向两边扩展（回文的中心点能够是摸个字符。也能够是某两个字符的中间），代码例如以下：

string expandAroundCenter(string s, int c1, int c2) {
  int l = c1, r = c2;
  int n = s.length();
  while (l >= 0 && r <= n-1 && s[l] == s[r]) {
    l--;
    r++;
  }
  return s.substr(l+1, r-l-1);
}
class Solution {
public:
    string longestPalindrome(string s) 
    {
      int n = s.length();
      if (n == 0) return "";
      string longest = s.substr(0, 1);  // a single char itself is a palindrome
      for (int i = 0; i < n-1; i++) {
        string p1 = expandAroundCenter(s, i, i);
        if (p1.length() > longest.length())
          longest = p1;
     
        string p2 = expandAroundCenter(s, i, i+1);
        if (p2.length() > longest.length())
          longest = p2;
      }
      return longest;
    }
};