求字符串中最长回文

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.

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Solutions：

1)将sreverse得到revS，然后求s和revS的最长公共字串（不是最长公共子序列）。复杂度为O(N^2)，空间也是。但还是总提示超时。在本机上测试500长度用例用时161ms。

class Solution {
private:	
	string lcstr(string str1, string str2) {
		string ret;
		int size=str1.size();
		int i=0, j=0;
		int lowLoc=0, highLoc=0;
	//	vector<vector<int> > c(size+1, vector<int>(size+1));
		int **c=new int*[size+1];
		for(i=0; i<=size; ++i) {
			c[i]=new int[size+1];
		}
		for(i=0; i<=str1.size(); ++i) {
			c[i][0] = 0;
		} 
		for(j=0; j<=str2.size(); ++j) {
			c[0][j] = 0;
		}
		int maxLength=0;
		for(i=1; i<=str1.size(); ++i) {
			for(j=1; j<=str2.size(); ++j) {
				if(str1[i-1] == str2[j-1]) {
					c[i][j] = c[i-1][j-1]+1;
				}
				else {					
					c[i][j] = 0;
				}
				if(c[i][j]>maxLength) {
					maxLength=c[i][j];
					lowLoc=i;
					highLoc=j;
				}
			}
		}
		for(i=0; i<=size; ++i) {
			delete [] c[i];
			c[i]=NULL;
		}
		delete []c;
		return str1.substr(lowLoc-maxLength, maxLength);
	}
public:
	string longestPalindrome(string s) {
		int size=s.size();
//		vector<vector<int> > c(size+1, vector<int>(size+1));
//		vector<vector<int> > b(size+1, vector<int>(size+1));
		string revS;
		for(int i=s.size()-1; i>=0; --i) {
			revS+=s[i];
		}
		return lcstr(s, revS);		 
	}
};

2）从某一个字符向两侧对称地扩展，直到不能对称为止。但要注意像“aa”这样的，所以每到一个字符要分别做奇偶扩展。时间复杂度为O(N^2)。

class Solution {
public:
	string longExpand(string s, int low, int high) {
		//int l=low, h=high;
		while(low >= 0 && high<s.size() && s[low]==s[high]) {
			--low;
			++high;
		}
	//	string ret;
		return s.substr(low+1, high-low-1);
	//	return ret;
	}
	string longestPalindrome(string s) {
		string ret;
		string odd, even;
		for(int i=0; i<s.size(); ++i) {
			odd=longExpand(s, i, i);
			even=longExpand(s, i, i+1);
			if(odd.size() > ret.size()) {
				ret=odd;
			}
			if(even.size() > ret.size()) {
				ret=even;
			}
		}
		return ret;
	}
};

Manacher’s Algorithm，O（n）的算法（看不懂）

class Solution{
public:
	string longestPalindrome(string s) {
		string T = preProcess(s);
		int n = T.length();
		int *P = new int[n];
		int C = 0, R = 0;
		int i=0;
		for (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 i
			while (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 (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);
	}
private:
	int min(int a, int b) {
		return a<b?a:b;
	}
	string 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;
	}
};