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.
给定一个包含大写字母和小写字母的字符串，找到通过这些字母构造成的最长的回文串。
在构造过程中，请注意区分大小写。比如 "Aa" 不能当做一个回文字符串。
注意:
假设字符串的长度不会超过 1010。
示例 1:
输入: "abccccdd" 输出: 7 解释: 我们可以构造的最长的回文串是"dccaccd", 它的长度是 7。
20ms
1 class Solution {
2 func longestPalindrome(_ s: String) -> Int {
3 //
4 var arr:[Int] = Array(repeating: 0,count: 128)
5 //遍历字符串
6 for char in s.characters
7 {
8 arr[char.toInt()] += 1
9 }
10 var ans:Int = 0
11 //遍历数组
12 for i in 0..<128
13 {
14 ans += arr[i] / 2 * 2
15 if ans % 2 == 0 && arr[i] % 2 == 1
16 {
17 ans += 1
18 }
19 }
20 return ans
21 }
22 }
23 //Character扩展代码
24 extension Character
25 {
26 func toInt() -> Int
27 {
28 var num:Int = Int()
29 for scalar in String(self).unicodeScalars
30 {
31 num = Int(scalar.value)
32 }
33 return num
34 }
35 }
28ms
1 class Solution {
2 func longestPalindrome(_ s: String) -> Int {
3 var map: [Bool] = Array(repeating: false, count: 128)
4 var length = 0
5 for char in s.unicodeScalars {
6 let value = Int(char.value)
7 map[value] = !map[value]
8 if !map[value] {
9 length += 2
10 }
11 }
12
13 if length < s.count {
14 length += 1
15 }
16
17 return length
18 }
19 }
28ms
1 class Solution {
2 struct CharTracker {
3 let c: Character
4 let n: Int
5 }
6 func longestPalindrome(_ s: String) -> Int {
7 var charTracker: [CharTracker] = [CharTracker]()
8 var pSum = 0
9 var firstSwitch: Bool = true
10 var cSet = [Character : Int]()
11 s.forEach {
12 cSet[$0, default: 0] += 1
13 }
14 for (_, n) in cSet {
15 if (n & 1) == 0 {
16 pSum += n
17 } else {
18 pSum += n - 1
19 if (firstSwitch) {
20 pSum += 1
21 firstSwitch = false
22 }
23 }
24 }
25 return pSum
26 }
27 }
32ms
1 class Solution {
2 func longestPalindrome(_ s: String) -> Int {
3 var frequencyTable = [Character:Int]()
4 for ch in s {
5 frequencyTable[ch] = (frequencyTable[ch] ?? 0) + 1
6 }
7
8 var count = 0
9 for (key,value) in frequencyTable {
10 if value%2 == 1 {
11 count += (value-1)
12 frequencyTable[key] = 1
13 } else {
14 count += value
15 frequencyTable[key] = 0
16 }
17 }
18
19 for (_,value) in frequencyTable {
20 if value%2 == 1 {
21 count += 1
22 break
23 }
24 }
25
26 return count
27 }
28 }
36ms
1 class Solution {
2 func longestPalindrome(_ s: String) -> Int {
3 var longestPalindrome = 0
4 var singleChar = 0
5
6 // check that the string is not empty
7 guard s.count > 0 else { return 0 }
8
9 // if there is only 1 character, return 1
10 if s.count == 1 {
11 return 1
12 }
13
14 // there can be 1 odd character included
15 // the rest can only be included in multiples of 2
16 // create a hashmap to keep track of the number of occurances
17 var characters = [Character: Int]()
18 for char in s {
19 if let count = characters[char] {
20 characters[char] = count + 1
21 } else {
22 characters[char] = 1
23 }
24 }
25
26 // if there‘s a 1 or modulo > 0, set a variable to 1
27 // loop through the characters and divide by 2
28 for key in characters.keys {
29 if characters[key]! % 2 != 0 {
30 singleChar = 1
31 }
32
33 longestPalindrome += characters[key]! / 2 * 2
34 // divide by 2 to eliminate remainders
35 }
36
37 longestPalindrome += singleChar
38
39 return longestPalindrome
40 }
41 }
原文地址：https://www.cnblogs.com/strengthen/p/9782912.html
时间： 2024-10-08 16:06:39