【LeetCode & 剑指offer刷题】发散思维题9：Shuffle an Array

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

Shuffle an Array

Shuffle a set of numbers without duplicates.

Example:

// Init an array with set 1, 2, and 3.

int[] nums = {1,2,3};

Solution solution = new Solution(nums);

 

// Shuffle the array [1,2,3] and return its result. Any permutation of [1,2,3] must equally likely to be returned.

solution.shuffle();

 

// Resets the array back to its original configuration [1,2,3].

solution.reset();

 

// Returns the random shuffling of array [1,2,3].

solution.shuffle();

---

C++

 

//方法：洗牌算法 Fisher–Yates shuffle algorithm

//O(n),O(n)

#include <cstdlib>

#include <algorithm>

class Solution

{

private:

    vector<int> original; //定义成员变量

    vector<int> array;

public:

    Solution(vector<int> nums) //?这里为什么不用引用

    {

        srand(time(nullptr)); // 以当前时间为随机生成器的种子，这里要加，不加速度很慢（猜想可前没能是如果rand()之有srand,则每次都会运行srand(1)比较耗时)

        original = nums; //这里为深拷贝

        array = nums;

    }

   

    /** Resets the array to its original configuration and return it. */

    vector<int> reset()

    {

        return original;

    }

   

    /** Returns a random shuffling of the array. */

    vector<int> shuffle()

    {

        int i,j;

        for( i = array.size()-1; i>0; i--) //从后往前扫描

        {

            j = rand() % (i+1); //产生0~i的随机数(！！注意要产生0~i的随机数，而不是0~i-1,因为要包换不交换的情况）

            swap(array[i], array[j]); //用当前数与随机选择的数进行交换

        }

        return array;

    }

};

/*方法二：用stl中shuffle函数，如shuffle(v.begin(), v.end());

实现说明：rand() % (i+1) 实际上不准确，因为生成的数对于多数 i 值不均匀分布。正确实现将实际上需要重新实现 C++11 std::uniform_distributtion

*/

/**

 * 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();

 */