Leetcode 398.随机数索引

原题链接：Leetcode 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.

Example 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 * 10⁴
• -2³¹ <= nums[i] <= 2³¹ - 1
• target is an integer from nums.
• At most 10⁴ calls will be made to pick.

用一个映射值和对应下标的数组的哈希表来存储信息，pick(i)实现从i对应的数组中返回随机值即可

class Solution {
    // <值, 下标数组>的映射
    unordered_map> pos;
public:
    Solution(vector &nums) {
        for (int i = 0; i < nums.size(); ++i) {
            pos[nums[i]].push_back(i);
        }
    }

    int pick(int target) {
        auto &indices = pos[target];
        return indices[rand() % indices.size()];
    }
};

• 时间复杂度：O(n)，开销在初始化
• 空间复杂度：O(n)，需要存储n个数的信息到哈希表