查找字符串的最大回文长度的两种算法： 第一中方法：
public class Solution { /** * @param s a string which consists of lowercase or uppercase letters * @return the length of the longest palindromes that can be built */ public int longestPalindrome(String s) { //最长长度 int k = 0; int i = s.length(); List list = new ArrayList(); Map map = new HashMap(); for (int j = 0; j < i; j++) { if (!list.contains(s.charAt(j))) { list.add(s.charAt(j)); //计算该字符出现了多少次 Integer b = 1; for (int a = j + 1; a < i; a++) { if (s.charAt(j) == s.charAt(a)) { b = b + 1; } } map.put(s.charAt(j), b); } } //还没计算只有一个元素的字符 boolean isF = true; Set set = map.keySet(); for (Object object : set) { int sum = Integer.valueOf(map.get(object).toString()); if (sum == 1 && isF) { k = k + 1; isF = false; } else if (sum > 1) { if (set.size() == 1) { k = k + sum; } else { //大于1的部分取偶数 k = k + (sum / 2) * 2; } } } return k; } }第二种方法：考虑到时间和空间复杂度，建议用第二种方法！！！
public class Solution { /** * @param s a string which consists of lowercase or uppercase letters * @return the length of the longest palindromes that can be built */ public int longestPalindrome(String s) { String str = s; int i = str.length(); char[] chars = new char[128]; for (int a = 0; a < i; a++) { chars[str.charAt(a)]++; } boolean isF = true; int k = 0; for (int b = 0; b < 127; b++) { if (chars[b] > 0) { if (chars[b] % 2 != 0) { //只允许出现一个奇数 if (isF) { k++; isF = false; } //如果该字符是只出现一次，要不就在上面算过了，要不就已经有奇数了，所以不能算 if (chars[b] != 1) { k = k + (chars[b] / 2) * 2; } } else if (chars[b] % 2 == 0) { k = k + chars[b]; } } } return k; } }