labuladong 题解
难度中等给你一个整数数组 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]
import copy class Solution: def __init__(self, nums: List[int]): self.org = nums self.rand = copy.deepcopy(nums) def reset(self) -> List[int]: return self.org def shuffle(self) -> List[int]: for i in range(len(self.rand)): p = random.randint(0,i) self.rand[p], self.rand[i] = self.rand[i],self.rand[p] return self.rand # Your Solution object will be instantiated and called as such: # obj = Solution(nums) # param_1 = obj.reset() # param_2 = obj.shuffle()
标签:rand,shuffle,nums,self,洗牌,Solution,384,数组 From: https://www.cnblogs.com/zle1992/p/17467581.html