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‎.idea/modules.xml

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‎.idea/workspace.xml

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‎[0409][Longest Palindrome]/[0409][Longest Palindrome].iml

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<?xml version="1.0" encoding="UTF-8"?>
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<module type="JAVA_MODULE" version="4">
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<component name="NewModuleRootManager" inherit-compiler-output="true">
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<exclude-output />
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<content url="file://$MODULE_DIR$">
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<sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
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</content>
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<orderEntry type="inheritedJdk" />
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<orderEntry type="sourceFolder" forTests="false" />
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<orderEntry type="module-library">
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<library name="JUnit4">
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<CLASSES>
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<root url="jar://$MAVEN_REPOSITORY$/junit/junit/4.12/junit-4.12.jar!/" />
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<root url="jar://$MAVEN_REPOSITORY$/org/hamcrest/hamcrest-core/1.3/hamcrest-core-1.3.jar!/" />
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</CLASSES>
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<JAVADOC />
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<SOURCES />
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</library>
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</orderEntry>
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</component>
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</module>

‎[0409][Longest Palindrome]/src/Main.java

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import org.junit.Assert;
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import org.junit.Test;
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/**
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* @author: wangjunchao(王俊超)
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* @time: 2019-07-03 04:15
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**/
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public class Main {
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@Test
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public void test1() {
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Solution solution = new Solution();
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Assert.assertEquals(7, solution.longestPalindrome("abccccdd"));
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}
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@Test
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public void test2() {
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Solution solution = new Solution();
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Assert.assertEquals(7, solution.longestPalindrome("aAaaaaaa"));
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}
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}

‎[0409][Longest Palindrome]/src/Solution.java

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/**
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* https://leetcode.com/problems/longest-palindrome/
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*
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* @author: wangjunchao(王俊超)
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* @time: 2019-07-03 04:08
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**/
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public class Solution {
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/**
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* <pre>
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* Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
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*
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* This is case sensitive, for example "Aa" is not considered a palindrome here.
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*
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* Note:
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* Assume the length of given string will not exceed 1,010.
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*
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* Example:
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*
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* Input:
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* "abccccdd"
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*
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* Output:
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* 7
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*
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* Explanation:
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* One longest palindrome that can be built is "dccaccd", whose length is 7.
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* 思路
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* 1、先记录每个字符出现的次数
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* 2、取每个字符出现的最大的偶数
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* 3、如果还有剩下的单个字符，只取其中的一个
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* </pre>
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*/
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public int longestPalindrome(String s) {
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if (s == null || s.length() < 1) {
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return 0;
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}
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int[] counter = new int[256];
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for (int i = 0; i < s.length(); i++) {
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char ch = s.charAt(i);
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counter[ch - 'A']++;
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}
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int result = 0;
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int left = 0;
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for (int value : counter) {
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result += 0xFFFFFFFE & value;
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left |= 0b1 & value;
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}
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return result + left;
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}
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}

‎[0409][Longest Palindrome]/src/Solution1.java

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/**
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* https://leetcode.com/problems/longest-palindrome/
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*
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* @author: wangjunchao(王俊超)
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* @time: 2019-07-03 04:08
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**/
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public class Solution1 {
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/**
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* <pre>
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* Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
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*
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* This is case sensitive, for example "Aa" is not considered a palindrome here.
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*
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* Note:
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* Assume the length of given string will not exceed 1,010.
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*
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* Example:
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*
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* Input:
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* "abccccdd"
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*
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* Output:
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* 7
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*
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* Explanation:
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* One longest palindrome that can be built is "dccaccd", whose length is 7.
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* 思路
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* 1、先记录每个字符出现的次数
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* 2、取每个字符出现的最大的偶数
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* 3、如果还有剩下的单个字符，只取其中的一个
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* </pre>
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*/
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public int longestPalindrome(String s) {
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if (s == null || s.length() < 1) {
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return 0;
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}
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int[] counter = new int[256];
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for (int i = 0; i < s.length(); i++) {
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char ch = s.charAt(i);
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counter[ch - 'A']++;
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}
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int result = 0;
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int left = 0;
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for (int value : counter) {
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if (value > 0) {
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if (value % 2 == 0) {
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result += value;
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} else {
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result += value - 1;
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left = 1;
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}
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}
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}
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return result + left;
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}
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}

‎[0409][Longest Palindrome]/src/Solution2.java

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/**
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* https://leetcode.com/problems/longest-palindrome/
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* @author: wangjunchao(王俊超)
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* @time: 2019-07-03 04:08
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**/
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public class Solution2 {
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/**
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* <pre>
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* Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
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*
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* This is case sensitive, for example "Aa" is not considered a palindrome here.
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*
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* Note:
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* Assume the length of given string will not exceed 1,010.
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*
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* Example:
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*
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* Input:
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* "abccccdd"
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*
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* Output:
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* 7
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*
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* Explanation:
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* One longest palindrome that can be built is "dccaccd", whose length is 7.
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* 思路
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* 1、先记录每个字符出现的次数
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* 2、取每个字符出现的最大的偶数
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* 3、如果还有剩下的单个字符，只取其中的一个
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* </pre>
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*/
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public int longestPalindrome(String s) {
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if (s == null || s.length() < 1) {
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return 0;
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}
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int[] counter = new int[26];
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// TIP: 这里是大小写不敏感的
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for (int i = 0; i < s.length(); i++) {
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char ch = s.charAt(i);
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if (ch >= 'a' && ch <= 'z') {
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counter[ch - 'a']++;
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} else {
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counter[ch - 'A']++;
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}
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}
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int result = 0;
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int left = 0;
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for (int value : counter) {
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if (value > 0) {
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if (value % 2 == 0) {
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result += value;
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// counter[i] = 0;
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} else {
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result += value - 1;
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// counter[i] = 1;
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// 记录还有剩下的1
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left = 1;
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}
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}
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}
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return result + left;
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}
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}

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