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 indexifrom nums wherenums[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.
方法一：哈希表
思路：
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
c++代码：
class Solution {
// <值, 下标数组>的映射
unordered_map<int, vector<int>> pos;
public:
Solution(vector<int> &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()];
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution* obj = new Solution(nums);
* int param_1 = obj->pick(target);
*/
复杂度分析：
- 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
