zoukankan      html  css  js  c++  java
  • [LeetCode] 409. Longest Palindrome 最长回文

    Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.

    This is case sensitive, for example "Aa" is not considered a palindrome here.

    Note:
    Assume the length of given string will not exceed 1,010.

    Example:

    Input:
    "abccccdd"
    
    Output:
    7
    
    Explanation:
    One longest palindrome that can be built is "dccaccd", whose length is 7.

    给一个含有大写和小写字母的字符串, 找出能用这些字母组成的最长的回文。大小写敏感,字符长度不超过1010。

    解法:由于字母可以随意组合,求出字符个数为偶数的字母,回文有两种形式,一种是左右完全对称,还有一种是以中间字符为中心,左右对称,统计出所有偶数个字符的出现总和,如果有奇数个字符的话,最后结果再加1。

    Java:

    public int longestPalindrome(String s) {
            if(s==null || s.length()==0) return 0;
            HashSet<Character> hs = new HashSet<Character>();
            int count = 0;
            for(int i=0; i<s.length(); i++){
                if(hs.contains(s.charAt(i))){
                    hs.remove(s.charAt(i));
                    count++;
                }else{
                    hs.add(s.charAt(i));
                }
            }
            if(!hs.isEmpty()) return count*2+1;
            return count*2;
    } 

    Python:

    class Solution(object):
        def longestPalindrome(self, s):
            """
            :type s: str
            :rtype: int
            """
            ans = odd = 0
            cnt = collections.Counter(s)
            for c in cnt:
                ans += cnt[c]
                if cnt[c] % 2 == 1:
                    ans -= 1
                    odd += 1
            return ans + (odd > 0)
    

    Python:

    class Solution(object):
        def longestPalindrome(self, s):
            map = {}
            for i in s:
                if i in map:
                    map[i] += 1
                else:
                    map[i] = 1
           
            odd, even = 0, 0
            for key in map:
                if map[key] % 2:
                    odd += 1
                else:
                    even += 1
                    
            return even + 1 if odd else even
    

    Python:

    import collections
    
    class Solution(object):
        def longestPalindrome(self, s):
            """
            :type s: str
            :rtype: int
            """
            odds = 0
            for k, v in collections.Counter(s).iteritems():
                odds += v & 1
            return len(s) - odds + int(odds > 0)
    
        def longestPalindrome2(self, s):
            """
            :type s: str
            :rtype: int
            """
            odd = sum(map(lambda x: x & 1, collections.Counter(s).values()))
            return len(s) - odd + int(odd > 0)
    

    C++:

    class Solution {
    public:
        int longestPalindrome(string s) {
            int odds = 0;
            for (auto c = 'A'; c <= 'z'; ++c) {
                odds += count(s.cbegin(), s.cend(), c) & 1;
            }
            return s.length() - odds + (odds > 0);
        }
    };
    

      

      

    类似题目:

    [LeetCode] 5. Longest Palindromic Substring 最长回文子串

    [LeetCode] 125. Valid Palindrome 有效回文

    [LeetCode] 9. Palindrome Number 验证回文数字

    [LeetCode] 516. Longest Palindromic Subsequence 最长回文子序列

      

    All LeetCode Questions List 题目汇总

  • 相关阅读:
    小小小康
    GC日志补充
    一次GC问题定位
    mycat1.5~1.6的一个bug
    [转] java Statement和PreparedStatement批量更新
    java 中的instanceof 运算符
    Java学习篇之数组方法
    iOS7适配的一点小技巧
    iOS 中正确切换摄像头&正确实现设置帧率的方式
    iOS 音量键事件监控响应
  • 原文地址:https://www.cnblogs.com/lightwindy/p/9012183.html
Copyright © 2011-2022 走看看