LeetCode 380 Insert Delete GetRandom O(1)

LeetCode

Implement the RandomizedSet class:

RandomizedSet() Initializes the RandomizedSet object. bool insert(int val) Inserts an item val into the set if not present. Returns true if the item was not present, false otherwise.
bool remove(int val) Removes an item val from the set if present. Returns true if the item was present, false otherwise. int getRandom() Returns a random element from the current set of elements (it’s guaranteed that at least one element exists when this method is called). Each element must have the same probability of being returned.

You must implement the functions of the class such that each function works in average O(1) time complexity.

Example 1:

Input
[“RandomizedSet”, “insert”, “remove”, “insert”, “getRandom”, “remove”, “insert”, “getRandom”]
[[], [1], [2], [2], [], [1], [2], []]
Output
[null, true, false, true, 2, true, false, 2]

Explanation
RandomizedSet randomizedSet = new RandomizedSet();
randomizedSet.insert(1); // Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomizedSet.remove(2); // Returns false as 2 does not exist in the set.
randomizedSet.insert(2); // Inserts 2 to the set, returns true. Set now contains [1,2].
randomizedSet.getRandom(); // getRandom() should return either 1 or 2 randomly.
randomizedSet.remove(1); // Removes 1 from the set, returns true. Set now contains [2].
randomizedSet.insert(2); // 2 was already in the set, so return false.
randomizedSet.getRandom(); // Since 2 is the only number in the set, getRandom() will always return 2.

Constraints:

-2<sup>31</sup> <= val <= 2<sup>31</sup> - 1 At most 2 * ``10<sup>5</sup> calls will be made to insert, remove, and getRandom.
There will be *at least one element in the data structure when getRandom is called.

算法思路：

Dict + List

注意事项：

1. 检查若删除最后一个元素发现问题，remove中删除key要放在最后，不能放中间

Python代码：

class RandomizedSet(TestCases):

    def __init__(self):
        self.nums = []
        self.key_to_index = {}

    def insert(self, val: int) -> bool:
        if val in self.key_to_index:
            return False
        self.nums.append(val)
        self.key_to_index[val] = len(self.nums) - 1
        return True

    def remove(self, val: int) -> bool:
        if val not in self.key_to_index:
            return False
        index = self.key_to_index[val]
        last_val = self.nums[len(self.nums) - 1]
        self.nums[index] = last_val
        self.key_to_index[last_val] = index
        self.key_to_index.pop(val)  # remember to put it last
        self.nums.pop()
        return True

    def getRandom(self) -> int:
        return self.nums[random.randint(0, len(self.nums) - 1)]

算法分析：

时间复杂度为O(1)，空间复杂度O(n).