You have a set which contains all positive integers [1, 2, 3, 4, 5, ...].
Implement the SmallestInfiniteSet class:
SmallestInfiniteSet() Initializes the SmallestInfiniteSet object to contain all positive integers.int popSmallest() Removes and returns the smallest integer contained in the infinite set.void addBack(int num) Adds a positive integer num back into the infinite set, if it is not already in the infinite set.
Example 1:
Input
["SmallestInfiniteSet", "addBack", "popSmallest", "popSmallest", "popSmallest", "addBack", "popSmallest", "popSmallest", "popSmallest"]
[[], [2], [], [], [], [1], [], [], []]
Output
[null, null, 1, 2, 3, null, 1, 4, 5]
Explanation
SmallestInfiniteSet smallestInfiniteSet = new SmallestInfiniteSet();
smallestInfiniteSet.addBack(2); // 2 is already in the set, so no change is made.
smallestInfiniteSet.popSmallest(); // return 1, since 1 is the smallest number, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 2, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 3, and remove it from the set.
smallestInfiniteSet.addBack(1); // 1 is added back to the set.
smallestInfiniteSet.popSmallest(); // return 1, since 1 was added back to the set and
// is the smallest number, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 4, and remove it from the set.
smallestInfiniteSet.popSmallest(); // return 5, and remove it from the set.
Constraints:
1 <= num <= 10001000 calls will be made in total to popSmallest and addBack.