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[Leetcode]398. Random Pick Index(C++)

题目描述

题目链接:398. Random Pick Index

Given an integer array nums 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.

Implement the Solution class:

  • Solution(int[] nums) Initializes the object with the array nums.
  • int pick(int target) Picks a random index i from nums where nums[i] == target. If there are multiple valid i’s, then each index should have an equal probability of returning.

例子

例子 1

Input:
["Solution", "pick", "pick", "pick"]
[[[1, 2, 3, 3, 3]], [3], [1], [3]]
Output:
[null, 4, 0, 2]
Explanation:
Solution solution = new Solution([1, 2, 3, 3, 3]);
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(1); // It should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.

Constraints

  • 1 <= nums.length <= 2 * 10^4
  • -2^31 <= nums[i] <= 2^31 - 1
  • target is an integer from nums.
  • At most 10^4 calls will be made to pick.

解题思路

这道题思路比较简单,由于要返回对应目标的下标,所以我们需要用一个哈希表来对对应元素的下标进行存储,这样查询的时候只需要常数时间。由于数组中有重复元素,因此应该用向量来存储下标,在返回时随机返回一个即可。代码如下:

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#include <unordered_map>
#include <vector>

class Solution {
public:
Solution(std::vector<int>& nums) {
for (int i = 0; i < nums.size(); ++i) {
hmap_[nums[i]].push_back(i);
}
}

int pick(int target) {
std::vector<int>& indices = hmap_[target];
return indices[rand() % indices.size()];
}

private:
std::unordered_map<int, std::vector<int>> hmap_;
};
  • 时间复杂度:
    • constructor: O(n)
    • pick: O(1)
  • 空间复杂度: O(n)

GitHub 代码同步地址: 398.RandomPickIndex.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions