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// 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();
这道题让我们给数组洗牌,也就是随机打乱顺序,那么由于之前那道题Linked List Random Node我们接触到了水塘抽样Reservoir Sampling的思想,这道题实际上这道题也是用类似的思路,我们遍历数组每个位置,每次都随机生成一个坐标位置,然后交换当前遍历位置和随机生成的坐标位置的数字,这样如果数组有n个数字,那么我们也随机交换了n组位置,从而达到了洗牌的目的,这里需要注意的是i + rand() % (res.size() - i)不能写成rand() % res.size(),虽然也能通过OJ,但是根据这个帖子的最后部分的概率图表,前面那种写法不是真正的随机分布,应该使用Knuth shuffle算法,感谢热心网友们的留言,参见代码如下:
class Solution {
public:
Solution(vector<int> nums): v(nums) {}
/** Resets the array to its original configuration and return it. */
vector<int> reset() {
return v;
}
/** Returns a random shuffling of the array. */
vector<int> shuffle() {
vector<int> res = v;
for (int i = 0; i < res.size(); ++i) {
int t = i + rand() % (res.size() - i);
swap(res[i], res[t]);
}
return res;
}
private:
vector<int> v;
};
Shuffle a set of numbers without duplicates.
Example:
这道题让我们给数组洗牌,也就是随机打乱顺序,那么由于之前那道题Linked List Random Node我们接触到了水塘抽样Reservoir Sampling的思想,这道题实际上这道题也是用类似的思路,我们遍历数组每个位置,每次都随机生成一个坐标位置,然后交换当前遍历位置和随机生成的坐标位置的数字,这样如果数组有n个数字,那么我们也随机交换了n组位置,从而达到了洗牌的目的,这里需要注意的是i + rand() % (res.size() - i)不能写成rand() % res.size(),虽然也能通过OJ,但是根据这个帖子的最后部分的概率图表,前面那种写法不是真正的随机分布,应该使用Knuth shuffle算法,感谢热心网友们的留言,参见代码如下:
类似题目:
Linked List Random Node
LeetCode All in One 题目讲解汇总(持续更新中...)
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