[leetcode] 398. Random Pick Index @ python

最新推荐文章于 2025-05-20 01:45:00 发布

原创最新推荐文章于 2025-05-20 01:45:00 发布 · 192 阅读

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本文介绍了一种在含重复元素的整数数组中，针对给定目标数字随机返回其索引位置的算法实现。该算法利用字典存储数组中每个数字的所有索引，通过随机选择实现目标数字索引的等概率返回，适用于大数据量场景。

原题

Given an array of integers with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.

Note:
The array size can be very large. Solution that uses too much extra space will not pass the judge.

Example:

int[] nums = new int[] {1,2,3,3,3};
Solution solution = new Solution(nums);

// pick(3) should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(3);

// pick(1) should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(1);

解法

使用字典, 键是nums中的数字, values是数字的索引列表, 然后使用random.choice从列表中随机选择一个index即可.

代码

class Solution:

    def __init__(self, nums: 'List[int]'):
        self.data = collections.defaultdict(list)
        for i, n in enumerate(nums):
            self.data[n].append(i)            

    def pick(self, target: 'int') -> 'int':
        return random.choice(self.data[target])

# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.pick(target)