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

题目描述

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：

#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