按字母順序排列數(shù)組后，我需要添加一個(gè) for 循環(huán)操作來(lái)搜索其中一個(gè)數(shù)組中的特定元素。在這種情況下，我需要搜索“Rogers. Steve”。我應(yīng)該如何進(jìn)行？預(yù)先感謝您的所有幫助！我已經(jīng)嘗試了幾個(gè)例子，但我無(wú)法得到正確的結(jié)果......String[] arrayString=new String[]{ "Administrator", "Vale.Vicky", "Lane.Lois", "Kent.Clark", "Wayne.Bruce", "Parker.Peter", "Rogers.Steve", "Luther.Lex", "Osborn.Harry", "Tony.Stark"};String searchString="Rogers.Steve";//Loop until the length of the array//Print the index of the string on an arraySystem.out.println("Rogers.Steve");我的課。package helloworldjavabook;public class HelloWorldjavabook { public static void main(String[] args) { String[] user = {"Administrator", "Vale.Vicky", "Lane.Lois", "Kent.Clark", "Wayne.Bruce", "Parker.Peter", "Rogers.Steve", "Luther.Lex", " Osborn.Harry", "Tony.Stark"}; orderedGuests(user); String[] password = {"PASSWORD1", "ILOVEBRUCE", "ILOVECLARK", " ILOVELOIS", "IAMGOTHAM", "IAMSPECTACULAR", "IAMAMERICA", "IRUNLEXCORP", "IAMGREENGOBLIN", " IAMINVINCIBLE"}; orderedGuests(password); String id[] = {"1", "2", "3", "4", "5", "6", "7", "8", "9", "10"}; orderedGuests(id); } private static void orderedGuests(String[] hotel) { Arrays.sort(hotel); System.out.println(Arrays.toString(hotel)); }}
2 回答
一只甜甜圈
TA貢獻(xiàn)1836條經(jīng)驗(yàn) 獲得超5個(gè)贊
public void search() {
String[] arrayString = new String[]{
"Administrator",
"Vale.Vicky",
"Lane.Lois",
"Kent.Clark",
"Wayne.Bruce",
"Parker.Peter",
"Rogers.Steve",
"Luther.Lex",
"Osborn.Harry",
"Tony.Stark"};
String searchString = "Rogers.Steve";
Arrays.sort(arrayString);
System.out.println(Arrays.binarySearch(arrayString, searchString));
}
這其中的關(guān)鍵部分是
Arrays.binarySearch(arrayString, searchString)
UYOU
TA貢獻(xiàn)1878條經(jīng)驗(yàn) 獲得超4個(gè)贊
我想你正在尋找這樣的東西
String[] arrayString=new String[]{
"Administrator",
"Vale.Vicky",
"Lane.Lois",
"Kent.Clark",
"Wayne.Bruce",
"Parker.Peter",
"Rogers.Steve",
"Luther.Lex",
"Osborn.Harry",
"Tony.Stark"};
String searchString="Rogers.Steve";
int position = 0;
for (int i = 0; i < arrayString.length; i++) {
if(arrayString[i] == searchString ){
position = i;
}
}
if( position > 0 ){
System.out.println("Position: " + position);
} else {
System.out.println(searchString + " not found" );
}
添加回答
舉報(bào)
0/150
提交
取消