[leetcode] 5. Longest Palindromic Substring (Medium)

原题链接

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找到并返回最长回路子串

思路：
解法一：
最简单的双重遍历，判断s[i]到s[j]是不是回串。
Runtime: 610 ms, faster than 6.39% of Java 慢的不行

class Solution {
        public String longestPalindrome(String s) {
            int len=s.length();
            for (int i = 0; i < len; i++) {
                int subNum = i + 1;
                int subLen = len - i;
                for (int j = 0; j < subNum; j++) {
                    String subStr = s.substring(j, j + subLen);
                    if (isPalindrome(subStr))
                        return subStr;
                }
            }
            return "";
        }

        public boolean isPalindrome(String s){
            int beg = 0, end = s.length() - 1;
            while(beg<end){
                if(s.charAt(beg)!=s.charAt(end))
                    return false;
                beg++;
                end--;
            }
            return true;
        }
    }

解法二：
遍历一次，以每一个s[i]为中心，计算。
Runtime: 4 ms, faster than 100.00% of Java

class Solution {
        int len = 0, maxLength = 0, start = 0;

        public String longestPalindrome(String s) {
            char[] arr = s.toCharArray();
            len = s.length();
            if (len <= 1)
                return s;
            for (int i = 0; i < len; i++) {
                i = helper(arr, i);
            }
            return s.substring(start, start + maxLength);
        }

        public int helper(char[] arr, int k) {
            int i = k - 1, j = k;
            while (j < len - 1 && arr[j] == arr[j + 1])
                j++;
            int nextCenter = j++;
            while (i >= 0 && j < len && arr[i] == arr[j]) {
                i--;
                j++;
            }
            if (j - i - 1 > maxLength) {
                maxLength = j - i - 1;
                start = i + 1;
            }
            return nextCenter;
        }

    }