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  • 5. 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.

    Example:

    Input: "babad"
    
    Output: "bab"
    
    Note: "aba" is also a valid answer.
    

    Example:

    Input: "cbbd"
    
    Output: "bb"
    写一下动态规划的思路,这个是leetcode上的动态规划思路

    Approach #3 (Dynamic Programming) [Accepted]

    To improve over the brute force solution, we first observe how we can avoid unnecessary re-computation while validating palindromes. Consider the case ''ababa''''ababa''. If we already knew that ''bab''''bab'' is a palindrome, it is obvious that ''ababa''''ababa'' must be a palindrome since the two left and right end letters are the same.

    We define P(i,j)P(i,j) as following:

    P(i,j)={true,if the substring SiSj is a palindromefalse,otherwise. P(i,j)={true,if the substring Si…Sj is a palindromefalse,otherwise. 

    Therefore,

    P(i, j) = ( P(i+1, j-1) ext{ and } S_i == S_j )P(i,j)=(P(i+1,j1) and Si​​==Sj​​)

    The base cases are:

    P(i, i) = trueP(i,i)=true

    P(i, i+1) = ( S_i == S_{i+1} )P(i,i+1)=(Si​​==Si+1​​)

    This yields a straight forward DP solution, which we first initialize the one and two letters palindromes, and work our way up finding all three letters palindromes, and so on...

    Complexity Analysis

    • Time complexity : O(n^2)O(n2​​). This gives us a runtime complexity of O(n^2)O(n2​​).

    • Space complexity : O(n^2)O(n2​​). It uses O(n^2)O(n2​​) space to store the table.

    这是以abade为例,得到的dp矩阵

    下面上代码:

        public String longestPalindrome(String s) {
            boolean[][] flag = new boolean[s.length()][s.length()];
            int maxlen = 0,start = 0;
            for(int i = 0;i < s.length(); i++){
            	flag[i][i] = true;
            	maxlen = 1;
            	start = i;
            }
            for(int i = 0;i < s.length()-1; i++)
            	if(s.charAt(i)==s.charAt(i+1)){
            		flag[i][i+1] = true;
            		maxlen = 2;
            		start = i;
            	}
            for(int len = 3; len<= s.length(); len++)
            	for(int i = 0;i < s.length()-len+1; i++){
            		int j = i+len-1;
            		if(s.charAt(i)==s.charAt(j)&&flag[i+1][j-1]==true){
            			flag[i][j] = true;
            			maxlen = len;
            			start = i;
            		}
            	}
            return s.substring(start, start+maxlen);
        }
    

      

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  • 原文地址:https://www.cnblogs.com/lxk2010012997/p/6261452.html
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