版权声明：本文由wikiwen撰写，转载请注明出处 https://blog.csdn.net/kk55guang2/article/details/86176464
【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();
*/