# 第三周ARTS总结
# Algorithm
43ms | 40.90% Run time
38.3MB | 49.35% Memory
public String longestPalindrome(String s) {
if (s.length() == 0) {
return "";
}
String longestPalindrome = s.substring(0, 1);
int maxLength = longestPalindrome.length();
if (s.length() == 1) {
return longestPalindrome;
}
// assume the length of longest palindromic is odd
for (int i = 1; i < s.length(); i++) {
int j = 1;
while (i - j >= 0 && i + j < s.length()) {
if (s.charAt(i - j) == s.charAt(i + j)) {
if (i - j == 0 || i + j == s.length() - 1) {
if (2 * j + 1 > maxLength) {
longestPalindrome = s.substring(i - j, i + j + 1);
maxLength = 2 * j + 1;
}
break;
}
j++;
} else {
if (2 * j - 1 > maxLength) {
longestPalindrome = s.substring(i - j + 1, i + j);
maxLength = 2 * j - 1;
}
break;
}
}
}
// assume the length of longest palindromic is even
for (int i = 1; i < s.length(); i++) {
int j = 1;
while (i - j >= 0 && i + j - 1 < s.length()) {
if (s.charAt(i - j) == s.charAt(i + j - 1)) {
if (i - j == 0 || i + j - 1 == s.length() - 1) {
if (2 * j > maxLength) {
longestPalindrome = s.substring(i - j, i + j);
maxLength = 2 * j;
}
break;
}
j++;
} else {
if (2 * j - 2 > maxLength) {
longestPalindrome = s.substring(i - j + 1, i + j - 1);
maxLength = 2 * j - 2;
}
break;
}
}
}
return longestPalindrome;
}
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# Review
# Tip
TextView如果设置了textColorHighlight，利用ClickableSpan之后点击文字背景会变色Activity的四种启动模式standard每次都新建一个实例对象singleTop如果在任务栈顶发现了相同的实例则重用，否则新建并压入栈顶singleTask如果在任务栈中发现了相同的实例，将其上面的任务终止并移除，重用该实例。否则新建实例并入栈singleInstance允许不同应用，进程线程等共用一个实例，无论从何应用调用该实例都重用
如果用了谷歌地图的SDK，并且target API≥28(Android 9.0)，需要在
<application>中将org.apache.http.legacy设置为false
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