LeetCode(5) Longest Palindromic Substring解题报告

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最长回文串

问题:

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

大意:

给定一个字符串s,找到s的最长回文子串。字符串的最大长度为1000.

示例:

Example:
Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.

Example:
Input: “cbbd”
Output: “bb”

解题方案:

这道题的解题思路有很多:
1. 求出每个子串，然后判断其是否为回文串。时间复杂度为O(n2)
2. 遍历字符串每个字符，从这个字符向两边校验是否相等。时间复杂度为O(n2)
3. Manacher(马拉车算法)，时间复杂度为O(n),个人觉得是个很有意思的算法，附上算法详解(自己也是新学的，所以就不分析了) http://blog.csdn.net/dyx404514/article/details/42061017

manacher算法java版(accepted)

    static class Manacher {
        private int[] len;//长度为i最长回文串的长度=len - 1,从左往右计算
        private String s;//原字符串
        private char[] t;//将字符串中间插入'#'后的字符数组。


        public Manacher(String s) {
            super();
            this.s = s;
        }


        /**
         * 预处理字符串，例如，s="abba" 那么t="$#a#b#b#a#@"
         * $和@是为了防止边界检测
         */
        private void preprocess() {
            t = new char[s.length() * 2 + 3];
            t[0] = '$';
            t[s.length() * 2 + 2] = '@';
            for (int i = 0; i < s.length(); i++) {
                t[2*i + 1] = '#';
                t[2*i + 2] = s.charAt(i);
            }
            t[s.length()*2 + 1] = '#';
        }

        public void process() {
            preprocess();

            len = new int[t.length];
            int center = 0, right = 0;
            for (int i = 1; i < t.length - 1; i++) {
                int mirror = 2*center - i;

                if (right > i) {
                    len[i] = Math.min(right - i, len[mirror]); 
                }

                while (t[i + (1 + len[i])] == t[i - (1 + len[i])]) {
                    len[i]++;
                }

                if (i + len[i] > right) {
                    center = i;
                    right = i + len[i];
                }
            }
        }

        public String longestPalindromicSubstring() {
            int length = 0;
            int center = 0;
            for (int i = 1; i < len.length - 1; i++) {
                if (len[i] > length) {
                    length = len[i];
                    center = i;
                }
            }
            return s.substring((center - 1 -length) / 2, (center - 1 +length) / 2);
        }

        public static void main(String[] args) {
            String s = "ababaca";
            Manacher manacher = new Manacher(s);
            manacher.process();
            String str = manacher.longestPalindromicSubstring();
            System.out.println(str);
        }
    }