一 字符串中的最大回文串(第5题)
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of sis 1000.
Example:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example:
Input: "cbbd" Output: "bb"
1. 我的解法(accepted): 中心扩展
思路: 回文即代表有中心，一次遍历中，对于每个位置上的数字求一下最大的回文串即可，初始处理剪枝，合并掉同样的元素，如xxxxaaaxxxx，先常量级别把a合并掉; 遍历时候再次剪枝，遍历过的有一个maxLen，如果即将遍历的元素最大可能回文长度都不可能超过maxLen，不再遍历。
1 public static void main(String[] args) { 2 String str = "ababababa"; 3 System.out.println(longestPalindrome(str)); 4 5 } 6 7 public static String longestPalindrome(String s) { 8 if (s == null) { 9 return null; 10 } 11 char[] chars = s.toCharArray(); 12 int length = chars.length; 13 int maxLen = 0; 14 String maxStr = ""; 15 16 for (int i = 0; i < length; i++) { 17 18 // cut branch 19 int possibleLength = getMaxPossibleLength(i, length); 20 if (possibleLength < maxLen) { 21 continue; 22 } 23 24 String maxStrTmp = getMaxStrByIndex(i, chars); 25 if (maxLen < maxStrTmp.length()) { 26 maxLen = maxStrTmp.length(); 27 maxStr = maxStrTmp; 28 } 29 } 30 return maxStr; 31 } 32 33 private static int getMaxPossibleLength(int index, int length) { 34 int head = 0; 35 int tail = length - 1; 36 if (index == head || index == tail) { 37 return 1; 38 } 39 int result1 = index - head; 40 int result2 = tail - index; 41 42 int min = result1 <= result2 ? result1 : result2; 43 return min * 2 + 1; 44 } 45 46 private static String getMaxStrByIndex(int index, char[] chars) { 47 StringBuilder sb = new StringBuilder(String.valueOf(chars[index])); 48 int length = chars.length; 49 int head = index - 1; 50 int tail = index + 1; 51 52 // middle deal 53 while (true) { 54 if (head >= 0 && chars[index] == chars[head]) { 55 sb.insert(0, String.valueOf(chars[head--])); 56 } else if (tail <= length - 1 && chars[index] == chars[tail]) { 57 sb.append(String.valueOf(chars[tail++])); 58 } else { 59 break; 60 } 61 } 62 63 // besides deal 64 while (true) { 65 if (head < 0 || tail > length - 1) { 66 break; 67 } 68 if (head >= 0 && tail <= length - 1 && chars[head] == chars[tail]) { 69 sb.insert(0, String.valueOf(chars[head--])); 70 sb.append(String.valueOf(chars[tail++])); 71 continue; 72 } 73 break; 74 75 } 76 return sb.toString(); 77 }
2. dp解法
