Longest Palindromic Substring 最長迴文字符串

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.

1、動態規劃

class Solution {
public:
    string longestPalindrome(string s) {
        if(s == "" || s.length()==0)
            return "";
        int len = s.length();
        
        bool flag[len][len];
        string res = "";
        int start = 0;
        int maxLen = 0;
        for(int i=len-1;i>=0;i--)
        {
            for(int j=i;j<len;j++)
            {
                if(s[i]==s[j] && (j-i<=2 || flag[i+1][j-1]))
                {
                    flag[i][j] = true;
                    if(maxLen<j-i+1)
                    {
                        maxLen=j-i+1;
                        start = i;
                        //res = s.substr(i,maxLen);
                    }
                }else{
                    flag[i][j] = false;
                }
            }
        }
        return s.substr(start,maxLen);
    }    
};

2、Manacher算法處理字符串迴文

class Solution {
public:
    string longestPalindrome(string s) {
		string t = "$";
		for (char ch : s) {
			t += '#';
			t += ch;
		}
		t += '#';
		int len = t.length();
		// t爲處理過的字符串，p爲記錄長度的數組 
        int p[t.length()];
		// mx爲已判斷迴文串最右邊位置，id爲中間位置，mmax記錄p數組中最大值  
		int mx = 0, id = 0, mmax = 0;  
		int right = 0;
		for (int i = 1; i < len; i++) {  
			p[i] = mx > i ? min(p[2 * id - i], mx - i) : 1;  
			while (t[i + p[i]] == t[i - p[i]])  
				p[i]++;  
			if (i + p[i] > mx) {  
				mx = i + p[i];  
				id = i;  
			}  
			if (mmax < p[i]) {
				mmax = p[i]; 
				right = i;
			}
		}  
		// 最長爲mmax - 1  
		return s.substr(right/2 - mmax/2, mmax-1);
    }
    
    
};

相關文章：http://blog.csdn.net/pi9nc/article/details/9251455