查找字符串的最大回文长度

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;
    }
}