【Leetcode】380. Insert Delete GetRandom O(1)
2017-02-25 16:00
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380. Insert Delete GetRandom O(1)
Design a data structure that supports all following operations in average O(1) time.
must have the same probability of being returned.
Example:
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题目简述:
对输入的数据实现,随机读取,插入,删除。。。并且复杂度是O(1)
思路简述:
看了网上的一些blog,大致的思路是 数组+哈希表。
把输入的数据存成两份,一份数组,一份哈希表。在哈希表上存的时候,key=输入的元素,value=元素在数组的索引.这样,删除的时候先去哈希表上找到元素在数组中的索引,然后再去数组中删除(ps:删除的时候,让数组最后一个元素与该元素对调然后删除最后一个元素),就实现了删除操作O(1)。同样,插入O(1)。利用数组特性,随机读取O(1);
代码:
import java.util.Vector;
public class RandomizedSet {
Map<Integer,Integer> aMap;
Vector<Integer> aVec;
//int index;
public RandomizedSet() {
aMap= new HashMap<Integer,Integer>();
aVec = new Vector<Integer>();
//index = 0;
}
public boolean insert(int val) {
int temp=aVec.size();
if(!aMap.containsKey(val)){
aMap.put(val,aVec.size());
aVec.add(val);
if(aVec.size()!=temp+1){
System.out.println("error");
}
return true;
}
return false;
}
public boolean remove(int val) {
if(aMap.containsKey(val)){
int temp = aMap.get(val);
aVec.set(temp,aVec.get(aVec.size()-1));
aMap.put(aVec.get(temp),temp);
aMap.remove(val);
aVec.remove(aVec.size()-1);
return true;
}
return false;
}
public int getRandom() {
Random rand = new Random();
//int num = ;
return aVec.get(rand.nextInt(aVec.size()));
}
}
运行结果:
Design a data structure that supports all following operations in average O(1) time.
insert(val): Inserts an item val to the set if not already present.
remove(val): Removes an item val from the set if present.
getRandom: Returns a random element from current set of elements. Each element
must have the same probability of being returned.
Example:
// Init an empty set. RandomizedSet randomSet = new RandomizedSet(); // Inserts 1 to the set. Returns true as 1 was inserted successfully. randomSet.insert(1); // Returns false as 2 does not exist in the set. randomSet.remove(2); // Inserts 2 to the set, returns true. Set now contains [1,2]. randomSet.insert(2); // getRandom should return either 1 or 2 randomly. randomSet.getRandom(); // Removes 1 from the set, returns true. Set now contains [2]. randomSet.remove(1); // 2 was already in the set, so return false. randomSet.insert(2); // Since 2 is the only number in the set, getRandom always return 2. randomSet.getRandom();
Subscribe to see which companies asked this question.
题目简述:
对输入的数据实现,随机读取,插入,删除。。。并且复杂度是O(1)
思路简述:
看了网上的一些blog,大致的思路是 数组+哈希表。
把输入的数据存成两份,一份数组,一份哈希表。在哈希表上存的时候,key=输入的元素,value=元素在数组的索引.这样,删除的时候先去哈希表上找到元素在数组中的索引,然后再去数组中删除(ps:删除的时候,让数组最后一个元素与该元素对调然后删除最后一个元素),就实现了删除操作O(1)。同样,插入O(1)。利用数组特性,随机读取O(1);
代码:
import java.util.Vector;
public class RandomizedSet {
Map<Integer,Integer> aMap;
Vector<Integer> aVec;
//int index;
public RandomizedSet() {
aMap= new HashMap<Integer,Integer>();
aVec = new Vector<Integer>();
//index = 0;
}
public boolean insert(int val) {
int temp=aVec.size();
if(!aMap.containsKey(val)){
aMap.put(val,aVec.size());
aVec.add(val);
if(aVec.size()!=temp+1){
System.out.println("error");
}
return true;
}
return false;
}
public boolean remove(int val) {
if(aMap.containsKey(val)){
int temp = aMap.get(val);
aVec.set(temp,aVec.get(aVec.size()-1));
aMap.put(aVec.get(temp),temp);
aMap.remove(val);
aVec.remove(aVec.size()-1);
return true;
}
return false;
}
public int getRandom() {
Random rand = new Random();
//int num = ;
return aVec.get(rand.nextInt(aVec.size()));
}
}
运行结果:
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