(LeetCode) 384. Shuffle an Array

384. Shuffle an Array

• Explore : Interview > Top Interveiw Questions > Easy Collection
• 분류 : Design
• 난이도 : Medium

Problem

Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.

Implement the Solution class:

• Solution(int[] nums) Initializes the object with the integer array nums.
• int[] reset() Resets the array to its original configuration and returns it.
• int[] shuffle() Returns a random shuffling of the array.

Example 1

Input
["Solution", "shuffle", "reset", "shuffle"]
[[[1, 2, 3]], [], [], []]
Output
[null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]

Explanation
Solution solution = new Solution([1, 2, 3]);
solution.shuffle();    // Shuffle the array [1,2,3] and return its result.
                       // Any permutation of [1,2,3] must be equally likely to be returned.
                       // Example: return [3, 1, 2]
solution.reset();      // Resets the array back to its original configuration [1,2,3]. Return [1, 2, 3]
solution.shuffle();    // Returns the random shuffling of array [1,2,3]. Example: return [1, 3, 2]

Constraints

• 1 <= nums.length <= 200
• -10^6 <= nums[i] <= 10^6
• All the elements of nums are unique.
• At most 5 * 10^4 calls in total will be made to reset and shuffle.

Solution

Exapnder

class Solution(nums: IntArray) {

    val array = nums
    
    /** Resets the array to its original configuration and return it. */
    fun reset(): IntArray {
        return array
    }

    /** Returns a random shuffling of the array. */
    fun shuffle(): IntArray {
        return array.clone().toList().shuffled().toIntArray()
    }

}

/**
 * Your Solution object will be instantiated and called as such:
 * var obj = Solution(nums)
 * var param_1 = obj.reset()
 * var param_2 = obj.shuffle()
 */

Point of Thinking

• 클래스 디자인 문제이다.
• 주어진 nums를 별도로 아카이브해두면 reset 메서드는 바로 처리할 수 있다.
• intArray를 shuffle하면 간단하게 끝날거라고 생각했는데, Array의 shuffle은 Unit을 반환한다.
• 셔플 이후 객체를 얻기 위해 List로 변환 후 shuffled를 호출한 뒤, 다시 intArray로 변환하면 Accepted.
• 아래 kotlin reference를 참고하자.

@SinceKotlin("1.4")
public fun IntArray.shuffle(random: Random): Unit {
    for (i in lastIndex downTo 1) {
        val j = random.nextInt(i + 1)
        val copy = this[i]
        this[i] = this[j]
        this[j] = copy
    }
}

/**
 * Returns a new list with the elements of this list randomly shuffled.
 */
@SinceKotlin("1.2")
public actual fun <T> Iterable<T>.shuffled(): List<T> = toMutableList().apply { shuffle() }

@SinceKotlin("1.2")
public actual fun <T> MutableList<T>.shuffle(): Unit {
    for (i in lastIndex downTo 1) {
        val j = Random.nextInt(i + 1)
        val copy = this[i]
        this[i] = this[j]
        this[j] = copy
    }
}