5 longest palindromic substring 最长回文子串

输入:s = “babad”
输出:“bab”

/*
    暴力破解法n^3
    依次判断是否是回文,并记录
*/
class Solution {
    
    
    public String longestPalindrome(String s) {
    
    
        char[] charArr = s.toCharArray();
        int len = charArr.length;
 
        //比较两个以上的数是否是回文
        int begin = 0, maxLen = 1;
        for(int i=0; i<len-1; i++)
        {
    
    
            for(int j=i+1; j<len; j++)
            {
    
    
                if( j-i+1 > maxLen && huiWen( charArr,i,j ) )
                {
    
    
                    maxLen = j-i+1;
                    begin = i;
                }
            }
        }
 
        return s.substring( begin, begin+maxLen );
    }
 
    //是否是回文
    public boolean huiWen( char[] arr, int begin, int end )
    {
    
    
        while( begin < end )
        {
    
    
            if( arr[begin] == arr[end] )
            {
    
    
                begin++;
                end--;
            }
            else
                return false;
        }
        return true;
    }
}
 
/*
    中心扩散法 n^2
    判断以中心扩散出去的是否是回文
        奇数:1个
        偶数:2个
*/
class Solution {
    
    
    public String longestPalindrome(String s) {
    
    
        char[] charArr = s.toCharArray();
        int len = charArr.length;
 
        //确保第二个不越界
        String res = s.substring(0,1);
        for(int i=0; i<len-1; i++)
        {
    
    
            String oddString = centerFun( charArr, i, i,s );
            String ouString = centerFun( charArr, i, i+1,s );
            String maxString = oddString.length() > ouString.length() ? oddString : ouString;
            if( maxString.length() > res.length() )
                res = maxString;
        }
 
        return res;
    }
 
    public String centerFun( char[] arr, int left, int right, String s )
    {
    
    
        while( left>=0 && right<arr.length )
        {
    
    
            if( arr[left] == arr[right] )
            {
    
    
                left--;
                right++;
            }
            else
                break;
        }
 
        return s.substring( left+1, right );
    }
}
 

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转载自blog.csdn.net/weixin_43891573/article/details/115196984