LeetCode 380. Insert Delete GetRandom O(1)
2017-08-27 15:10
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问题描述:
Design a data structure that supports all following operations in average O(1) time.
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();
问题描述:设计一个类,其中插入,删除,随机值都是O(1)的复杂度。
随机插入,先检查值是否存在。需要实现O(1)所以必须是Hash 来实现。删除也一样。
第三个条件随机一个数,所有的数随机到的概率是相同的。数需要被存在在Array中。
所以使用一个ArrayList来存放value,然后使用一个HashMap来存放value-index的map。
这就带来了一个新的问题,删除一个数,需要做到O(1).正常的删除需要移位,复杂度是O(n).这里有一个小技巧:将array尾部的值换到被删除value的位置上,然后删除尾部的值。
代码如下:
class RandomizedSet {
private Map<Integer, Integer> valueIndexMap;
private List<Integer> valueList;
private Random random = new Random();
/** Initialize your data structure here. */
public RandomizedSet() {
valueIndexMap = new HashMap<Integer, Integer>();
valueList = new ArrayList<Integer>();
}
public boolean insert(int val) {
if (valueIndexMap.containsKey(val)) {
return false;
}
int index = valueList.size();
valueList.add(index, val);
valueIndexMap.put(val, index);
return true;
}
/**
* 删除一个元素,正常的做法需要移动位置,可以固定的删除最后一位,然后将最后一位和删除的额位置调换
*/
public boolean remove(int val) {
if (!valueIndexMap.containsKey(val)) {
return false;
}
Integer index = valueIndexMap.get(val);
int lastIndex = valueList.size() - 1;
if (index != (lastIndex)) {
Integer lastValue = valueList.get(lastIndex);
valueList.set(index, lastValue);
valueIndexMap.put(lastValue, index);// 修改最后一位的数值
}
valueList.remove(lastIndex);
valueIndexMap.remove(val);
return true;
}
/**
* 下一个随机数
*/
public int getRandom() {
return valueList.get(random.nextInt(valueList.size()));
}
}
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();
问题描述:设计一个类,其中插入,删除,随机值都是O(1)的复杂度。
随机插入,先检查值是否存在。需要实现O(1)所以必须是Hash 来实现。删除也一样。
第三个条件随机一个数,所有的数随机到的概率是相同的。数需要被存在在Array中。
所以使用一个ArrayList来存放value,然后使用一个HashMap来存放value-index的map。
这就带来了一个新的问题,删除一个数,需要做到O(1).正常的删除需要移位,复杂度是O(n).这里有一个小技巧:将array尾部的值换到被删除value的位置上,然后删除尾部的值。
代码如下:
class RandomizedSet {
private Map<Integer, Integer> valueIndexMap;
private List<Integer> valueList;
private Random random = new Random();
/** Initialize your data structure here. */
public RandomizedSet() {
valueIndexMap = new HashMap<Integer, Integer>();
valueList = new ArrayList<Integer>();
}
public boolean insert(int val) {
if (valueIndexMap.containsKey(val)) {
return false;
}
int index = valueList.size();
valueList.add(index, val);
valueIndexMap.put(val, index);
return true;
}
/**
* 删除一个元素,正常的做法需要移动位置,可以固定的删除最后一位,然后将最后一位和删除的额位置调换
*/
public boolean remove(int val) {
if (!valueIndexMap.containsKey(val)) {
return false;
}
Integer index = valueIndexMap.get(val);
int lastIndex = valueList.size() - 1;
if (index != (lastIndex)) {
Integer lastValue = valueList.get(lastIndex);
valueList.set(index, lastValue);
valueIndexMap.put(lastValue, index);// 修改最后一位的数值
}
valueList.remove(lastIndex);
valueIndexMap.remove(val);
return true;
}
/**
* 下一个随机数
*/
public int getRandom() {
return valueList.get(random.nextInt(valueList.size()));
}
}
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