题目要求
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.
输入一个字符串,计算用这个字符串中的值构成一个最长回数的长度是多少。
思路和代码
这是一道 easy 难度的题目,但是一次性写对也有挑战。直观来看,我们立刻就能想到统计字符串中每个字符出现的次数,如果该字符出现次数为偶数,则字符一定存在于回数中。但是我们忽略了一点,即如果字符中存在一个额外的单个字符位于中间,该字符串也能构成回数,如 aabaa。这个细节需要注意。下面是 O(N) 时间的实现:
public int longestPalindrome(String s) {
int[] count = new int[52];
int max = 0;
for(char c : s.toCharArray()) {
if(c>=’a’ && c<=’z’){
count[c-‘a’]++;
if(count[c-‘a’] % 2 == 0) {
max +=2;
}
}
if(c>=’A’ && c<=’Z’){
count[c-‘A’ + 26]++;
if(count[c-‘A’+26] % 2 == 0) {
max += 2;
}
}
}
if(max < s.length()) {
max++;
}
return max;
}