算法练习——最长回文子串

输入: "babad"
输出: "bab"
注意: "aba"也是一个有效答案。

输入: "cbbd"
输出: "bb"

public class LongestPalindrome1 {
    
    //利用两个for循环来遍历所有的子串，并且在里面再加一层while循环用于判断字符串是否是回文串
    public String longestPalindrome(String s) {
        for (int size = s.length(); size > 0; size--) {
            for (int low = 0, hight = low+size-1; hight < s.length(); low++,hight++) {
                if (isPalindrome(s,low,hight)) {
                    return s.substring(low, hight+1);
                }
            }
        }
        return s.substring(0, 1);
    }
    
    //判断一个字符串是否是回文字符
    private boolean isPalindrome(String s, int low, int hight) {
        while (low < hight) {
            if(s.charAt(low) == s.charAt(hight)){
                low++;
                hight--;
            }else {
                return false;
            }
        }
        return true;
    }
}

public class LongestPalindrome {
    public static String longestPalSubstr(String s){
        
        int n = s.length();
        
        boolean[][] table = new boolean[n][n];
        
        //所有的一个字符肯定是回文串
        int maxLength = 1;
        
        //组成的boolean矩阵的对角都肯定为true
        for (int i = 0; i < n; i++) {
            table[i][i] = true;
        }
        
        int start = 0;
        
        //如果字符串长度小于2，则直接处理
        for (int i = 0; i < n - 1; ++i) {
            if (s.charAt(i) == s.charAt(i +1)) {
                table[i][i+1] = true;
                start = i;
                maxLength = 2;
            }
        }
        
        //如果字符串长度大于2，则开始遍历处理，主要是通过两层for循环
        
        //检查字符串长度大于2的字符串，其中k表示子字符串的长度
        for (int k = 3; k <= n; ++k) {
            
            //i表示子字符串中起始的位置
            for (int i = 0; i < n - k + 1; i++) {
                
                //获取子字符串结束的位置，
                int j = i + k - 1;
                
                if (s.charAt(i) == s.charAt(j) && table[i+1][j-1]) {
                    table[i][j] = true;
                    if (k > maxLength) {
                        maxLength = k;
                        start = i;
                    }
                }
            }
        }
        return s.substring(start, start+maxLength);
    }    
}

public class LongestPalindrome {

    public static String longestPalSubstr(String s){
        //最后字符串的长度
        int maxLength = 1;  
        
        int start = 0;
        int len  = s.length();
        
        int low ,hight;
        
        
        //这里面用i表示中心点的位置
        for (int i = 0; i < len; i++) {
            
            //首先查找最长的偶数回文字符串,有两个中心点，分别为 i -1、 i
            low = i -1;
            hight = i;
            while (low >= 0 && hight < len && s.charAt(low) == s.charAt(hight)) {
                if (hight - low + 1 > maxLength) {
                    maxLength = hight - len + 1;
                    start = low;
                }
                --low;
                ++hight;
            }
            
            //其次寻找最长的奇数的最长回文字符串，中心点以i为主
            low = i - 1;
            hight = i + 1;
            while (low >= 0 && hight < len && s.charAt(low) == s.charAt(hight)) {
                if (hight - low + 1 > maxLength) {
                    maxLength = hight - low + 1;
                    start = low;
                }
                --low;
                ++hight;
            }
        }
        
        //起始和结束的位置
        return s.substring(start, start + maxLength);
    }
}