384. 打乱数组

题目描述

给你一个整数数组 nums ，设计算法来打乱一个没有重复元素的数组。打乱后，数组的所有排列应该是 等可能 的。

实现 Solution class:

• Solution(int[] nums) 使用整数数组 nums 初始化对象
• int[] reset() 重设数组到它的初始状态并返回
• int[] shuffle() 返回数组随机打乱后的结果

 

示例 1：

输入
["Solution", "shuffle", "reset", "shuffle"]
[[[1, 2, 3]], [], [], []]
输出
[null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]

解释
Solution solution = new Solution([1, 2, 3]);
solution.shuffle();    // 打乱数组 [1,2,3] 并返回结果。任何 [1,2,3]的排列返回的概率应该相同。例如，返回 [3, 1, 2]
solution.reset();      // 重设数组到它的初始状态 [1, 2, 3] 。返回 [1, 2, 3]
solution.shuffle();    // 随机返回数组 [1, 2, 3] 打乱后的结果。例如，返回 [1, 3, 2]

 

提示：

• 1 <= nums.length <= 50
• -106 <= nums[i] <= 106
• nums 中的所有元素都是 唯一的
• 最多可以调用 104 次 reset 和 shuffle

解法

方法一

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def __init__(self, nums: List[int]):
        self.nums = nums
        self.original = nums.copy()

    def reset(self) -> List[int]:
        self.nums = self.original.copy()
        return self.nums

    def shuffle(self) -> List[int]:
        for i in range(len(self.nums)):
            j = random.randrange(i, len(self.nums))
            self.nums[i], self.nums[j] = self.nums[j], self.nums[i]
        return self.nums


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

class Solution {
    private int[] nums;
    private int[] original;
    private Random rand;

    public Solution(int[] nums) {
        this.nums = nums;
        this.original = Arrays.copyOf(nums, nums.length);
        this.rand = new Random();
    }

    public int[] reset() {
        nums = Arrays.copyOf(original, original.length);
        return nums;
    }

    public int[] shuffle() {
        for (int i = 0; i < nums.length; ++i) {
            swap(i, i + rand.nextInt(nums.length - i));
        }
        return nums;
    }

    private void swap(int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(nums);
 * int[] param_1 = obj.reset();
 * int[] param_2 = obj.shuffle();
 */

class Solution {
public:
    vector<int> nums;
    vector<int> original;

    Solution(vector<int>& nums) {
        this->nums = nums;
        this->original.resize(nums.size());
        copy(nums.begin(), nums.end(), original.begin());
    }

    vector<int> reset() {
        copy(original.begin(), original.end(), nums.begin());
        return nums;
    }

    vector<int> shuffle() {
        for (int i = 0; i < nums.size(); ++i) {
            int j = i + rand() % (nums.size() - i);
            swap(nums[i], nums[j]);
        }
        return nums;
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution* obj = new Solution(nums);
 * vector<int> param_1 = obj->reset();
 * vector<int> param_2 = obj->shuffle();
 */

type Solution struct {
    nums, original []int
}

func Constructor(nums []int) Solution {
    return Solution{nums, append([]int(nil), nums...)}
}

func (this *Solution) Reset() []int {
    copy(this.nums, this.original)
    return this.nums
}

func (this *Solution) Shuffle() []int {
    n := len(this.nums)
    for i := range this.nums {
        j := i + rand.Intn(n-i)
        this.nums[i], this.nums[j] = this.nums[j], this.nums[i]
    }
    return this.nums
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(nums);
 * param_1 := obj.Reset();
 * param_2 := obj.Shuffle();
 */

class Solution {
    private nums: number[];

    constructor(nums: number[]) {
        this.nums = nums;
    }

    reset(): number[] {
        return this.nums;
    }

    shuffle(): number[] {
        const n = this.nums.length;
        const res = [...this.nums];
        for (let i = 0; i < n; i++) {
            const j = Math.floor(Math.random() * n);
            [res[i], res[j]] = [res[j], res[i]];
        }
        return res;
    }
}

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

use rand::Rng;
struct Solution {
    nums: Vec<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl Solution {
    fn new(nums: Vec<i32>) -> Self {
        Self { nums }
    }

    fn reset(&self) -> Vec<i32> {
        self.nums.clone()
    }

    fn shuffle(&mut self) -> Vec<i32> {
        let n = self.nums.len();
        let mut res = self.nums.clone();
        for i in 0..n {
            let j = rand::thread_rng().gen_range(0, n);
            res.swap(i, j);
        }
        res
    }
}/**
 * Your Solution object will be instantiated and called as such:
 * let obj = Solution::new(nums);
 * let ret_1: Vec<i32> = obj.reset();
 * let ret_2: Vec<i32> = obj.shuffle();
 */

/**
 * @param {number[]} nums
 */
const Solution = function (nums) {
    this.nums = nums || [];
};

/**
 * Resets the array to its original configuration and return it.
 * @return {number[]}
 */
Solution.prototype.reset = function () {
    return this.nums;
};

/**
 * Returns a random shuffling of the array.
 * @return {number[]}
 */
Solution.prototype.shuffle = function () {
    let a = this.nums.slice();
    for (let i = 0; i < a.length; i++) {
        let rand = Math.floor(Math.random() * (a.length - i)) + i;
        let tmp = a[i];
        a[i] = a[rand];
        a[rand] = tmp;
    }
    return a;
};

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

评论