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[Day10]5. Longest Palindromic Su

[Day10]5. Longest Palindromic Su

作者: Shira0905 | 来源:发表于2017-02-08 23:35 被阅读0次

    DESCRIPTION:

    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"

    ANALYSIS:

    When I first saw this problem, some wrong ideas up to me. So, I asked SLF for some hints.
    Then, I followed his thought and code for almost one hour and finally AC.

    1. Loop 'i' is the index of the middle of palindromic substring, and expand to left and right one by one at the same time
    2. One point should be pay attention to is that, the middle index of palindromic substring can be .5. For example, 'abba', its middle index is 1.5.
    3. I am disappointed in my speed in solve the index range of 'l' and 'r'. It takes me a lot of time to implement the code without some bugs like indexOutOfBoundException. Maybe it reminds me to practice more.

    SOLUTION:

    public static String longestPalindrome(String s) {
        String longestString="";
        for(double i=0.5;i<s.length()-1;i=i+1){
            double l=i-0.5;
            double r=i+0.5;
            while(l>=0&&r<s.length()&&s.charAt((int) l)==s.charAt((int) r)){
                l--;r++;
            }
            l++;r--;
            String string=s.substring((int)l,(int)r+1);
            if(string.length()>longestString.length())
                longestString=string;
        }
        for(int i=0;i<s.length();i++){
            int l=i;
            int r=i;
            while (l>=0&&r<s.length()&&s.charAt(l)==s.charAt(r)) {
                l--;r++;
            }
            l++;r--;
            String string=s.substring(l,r+1);
            if(string.length()>longestString.length())
                longestString=string;
        }
        return longestString;
    }

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