分析題:
1.首先回文,就是前面讀 和 後面讀一樣 由此可見開頭和結尾是一致的
比如:abcba 前面abc 後面abc
2. 分析這道題的時候有以下情況,既然是前後讀一樣,它需要一個界點分前後,我就把字符串定爲:
1.偶數
2.奇數
3.不是迴文的情況
4.全是一個字母的情況
5.空串
開始寫代碼 ,經過了很多次代碼的修改
結果:超時了
然後發現如下代碼有如下的優化點:
-
代碼尋找前後字母相同 組裝成了一個空間複雜度的HashSet 然後在分析這個作用域中的值 是不是符合前,後讀相同
-
判斷是否是迴文的時候:
for (int i = count - 1; i >= midst; i–) {
stbAft.append(val.charAt(i));
}
append 會每一次都copy 一次數組
public class Solution {
public static String longestPalindrome(String val){
//判斷是否爲空
if (val == null || val.isEmpty() || val.length() == 1) {
return val;
}
//存儲所有開始和結尾相同 的String
HashSet<String> tempHashResult = new HashSet<>();
//查詢所有迴文
char[] charts = val.toCharArray();
//如果就一個字母直接return val
if (val.matches("["+charts[0]+"]{"+charts.length+"}")) {
return val;
}
for (int i = 0; i < charts.length; i++) {
char aVal = charts[i];
for (int j = i + 1; j < charts.length; j++) {
char bVal = charts[j];
if (aVal == bVal) {
tempHashResult.add(val.substring(i, j + 1));
}
}
}
int count = 0;
String result = "";
for (String cVal : tempHashResult) {
if ((cVal.length() - 2 == 1 || cVal.length() - 2 == 0 || confirmListLongestPalindrome(cVal)) && count < cVal.length()) {
count = cVal.length();
result = cVal;
}
}
result = result.length() == 0 ? val.substring(0, 1) : result;
//排序迴文字段
return result;
}
/**
* 判斷是否是迴文
*
* @param val 值
* @return
*/
private static boolean confirmListLongestPalindrome(String val) {
int count = val.length();
boolean isEvent = count % 2 == 0;
StringBuilder stbAft = new StringBuilder();
int midst = (int) Math.ceil(count / 2.0);
for (int i = count - 1; i >= midst; i--) {
stbAft.append(val.charAt(i));
}
return val.substring(0,isEvent ? midst : midst - 1).equals(stbAft.toString());
}
public static void main(String[] args) {
String val = "abababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababababa";
long ms = System.currentTimeMillis();
String bbb = longestPalindrome(val);
System.out.println("bbb = " + bbb);
long msA = System.currentTimeMillis();
System.out.println("Simple = " + (msA - ms));
}
}
最後優化上面的兩個點
import java.util.HashSet;
public class Main {
public static String longestPalindrome(String val) {
//判斷是否爲空
if (val == null || val.isEmpty() || val.length() == 1) {
return val;
}
//查詢所有迴文
char[] charts = val.toCharArray();
//如果就一個字母直接return val
if (val.matches("["+charts[0]+"]{"+charts.length+"}")) {
return val;
}
int count = 0;
String result = "";
for (int i = 0; i < charts.length; i++) {
char aVal = charts[i];
for (int j = i + 1; j < charts.length; j++) {
char bVal = charts[j];
if (aVal == bVal){
String cval = val.substring(i, j + 1);
if(confirmListLongestPalindrome(cval) && count < cval.length()) {
count = cval.length();
result = cval;
}
}
}
}
result = result.length() == 0 ? val.substring(0, 1) : result;
//排序迴文字段
return result;
}
/**
* 遞歸求值
*
* @param val 值
* @return
*/
private static boolean confirmListLongestPalindrome(String val) {
int count = val.length();
boolean isEvent = count % 2 == 0;
int midst = (int) Math.ceil(count / 2.0);
StringBuilder stbAft = new StringBuilder(val.substring(midst));
return val.substring(0, isEvent ? midst : midst - 1).equals(stbAft.reverse().toString());
}
/**
* aaaa
* adc
*
* @param args
*/
public static void main(String[] args) {
String val = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabcaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
long ms = System.currentTimeMillis();
String bbb = longestPalindrome(val);
System.out.println("bbb = " + bbb);
long msA = System.currentTimeMillis();
System.out.println("Simple = " + (msA - ms));
}
}