leetCode 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 1:

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

Example 2:

Input: "cbbd"
Output: "bb"

class Solution {
    public boolean isPalindromic(String s, int begin, int end) {
        if (begin == end)
            return true;
        for (int i = 0; i <= (end - begin) / 2; i++) {
            if (s.charAt(begin + i) != s.charAt(end - i))
                return false;
        }
        return true;
    }

    public String longestPalindrome(String s) {
        if (s.length() <= 1) return s;
        int[] res = new int[s.length()];// 存放以当前位置结束的最长回文子串的长度

        res[0] = 1;
        for (int i = 1; i < s.length(); i++) {
            int begin = i - res[i - 1] - 1;//以i-1位置结尾的最长回文子串的开始位置的前一个位置
            if (begin < 0) {
                /*
                 * 查找以i结尾的最长回文子串
                 * */
                for (int j = 0; j <= i; j++) {
                    if (isPalindromic(s, j, i)) {
                        res[i] = i - j + 1;
                        break;
                    }
                }
            } else {
                if (s.charAt(begin) == s.charAt(i)) {
                    res[i] = res[i - 1] + 2;
                } else {
                    // 判断begin+1到i子串是否是回文子串
                    for (int j = 0; j <= i; j++) {
                        if (isPalindromic(s, j, i)) {
                            res[i] = i - j + 1;
                            break;
                        }
                    }
                }
            }
        }
        int max = 0;
        for (int i = 0; i < s.length(); i++) {
            if (res[max] < res[i]) max = i;
        }
        return s.substring(max - res[max] + 1, max + 1);
    }
}

动态规划解法：d[i][j]：表示子串(i,j)是否是回文子串

class Solution {
    public String longestPalindrome(String s) {
        if (s.length() <= 0) return s;
        boolean[][] d = new boolean[s.length()][s.length()];
        int max = 0, begin = 0, end = 0;
        for (int i = 0; i < s.length(); i++) {
            d[i][i] = true;
        }
        for (int j = 1; j < s.length(); j++) {
            for (int i = 0; i < j; i++) {
                if (i == j - 1) {
                    if (s.charAt(i) == s.charAt(j)) {
                        d[i][j] = true;
                        if (max < j - i) {
                            begin = i;
                            end = j;
                            max = j - i;
                        }
                    } else {
                        d[i][j] = false;
                    }
                } else {
                    if (d[i + 1][j - 1]) {
                        if (s.charAt(i) == s.charAt(j)) {
                            d[i][j] = true;
                            if (max < j - i) {
                                begin = i;
                                end = j;
                                max = j - i;
                            }
                        } else {
                            d[i][j] = false;
                        }
                    } else {
                        d[i][j] = false;
                    }
                }
            }
        }
        return s.substring(begin, end+1);
    }
}

 第三种解法：由中间向两遍扩散查找回文串

class Solution {
    public String longestPalindrome(String s) {
        if (s.length() <= 1) return s;
        int max = 0;
        int low = 0, high = 0;
        //计算奇数回文子串的长度
        for (int i = 0; i < s.length(); i++) {
            int begin = i, end = i;
            while (begin > 0 && end < s.length() - 1) {
                --begin;
                ++end;
                if (s.charAt(begin) == s.charAt(end)) {
                    if (max < end - begin) {
                        max = end - begin;
                        low = begin;
                        high = end;
                    }
                } else {
                    break;
                }
            }
        }

        //计算偶数回文子串的长度
        for (int i = 0; i < s.length() - 1; i++) {
            int begin = i, end = i + 1;
            if (s.charAt(i) != s.charAt(i+1)) continue;
            else {
                if (max < end-begin){
                    max=end-begin;
                    low=begin;
                    high=end;
                }
            }
            while (begin > 0 && end < s.length() - 1) {
                --begin;
                ++end;
                if (s.charAt(begin) == s.charAt(end)){
                    if (max < end-begin){
                        max=end-begin;
                        low= begin;
                        high=end;
                    }
                }else {
                    break;
                }
            }
        }
        return s.substring(low, high + 1);
    }
}