QUESTION:
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 <= 1000
At most 1000 calls will be made in total to popSmallest and addBack.
EXPLANATION:
根据限制可以知道, num只有1000个, 那么我们就可以确定数据结构. 用array来进行. 如果pop出去了, 那么就改为false, 然后返回第一个为true的就可以. index就是对应的数字即可.
SOLUTION:
class SmallestInfiniteSet() {
var arr:BooleanArray = BooleanArray(1001 ){ i -> true}
init {
arr[0] = false
}
fun popSmallest(): Int {
var tmp:Int = 0
while (!arr[tmp]) {
tmp++
}
arr[tmp] = false
return tmp
}
fun addBack(num: Int) {
arr[num] = true
}
}