[LeetCode] (medium) 398. Random Pick Index

https://leetcode.com/problems/random-pick-index/

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);

---

直觀做法：

用map記錄每個數字和他出現的所有下標，然後用rand找

class Solution {
public:
    default_random_engine E;
    unordered_map<int, vector<int>> M;
        
    Solution(vector<int>& nums) {
        for(int i = 0; i < nums.size(); ++i){
            M[nums[i]].push_back(i);
        }
    }
    
    int pick(int target) {
        uniform_int_distribution<int> U(0, M[target].size()-1);
        return M[target][U(E)];
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution* obj = new Solution(nums);
 * int param_1 = obj->pick(target);
 */

 

更好的做法：

隨機尋找的方法，可證明：

1/k * (1-1/k+1)) * (1-1/k+2)) * ... * (1-1/n)

= 1/k * k/(k+1) * (k+1)/(k+2) * ... * (n-1)/n

= 1/n

class Solution {
    
public:
    vector<int> V;
    int len;
    
    Solution(vector<int>& nums):V(nums){
        len = nums.size();
        static int fast_io = []() { std::ios::sync_with_stdio(false); cin.tie(nullptr); return 0; }();
    }
    
    int pick(int target) {
        int count = 0, result = 0;
        
        for(int i = 0; i < len; ++i){
            if(V[i] == target){
                ++count;
                if(rand()%count == 0) result = i;
            }
        }
        
        return result;
    }
};

 

關於隨機性驗證我寫了個小的python腳本：

import random

def get_random():
    # print(int(random.random()*65536))
    return int(random.random()*65536)


test_cnt = int(input("how much times: "))
num_cnt = int(input("how many numbers: "))

cnt_list = []

for i in range(0, num_cnt):
    cnt_list.append(0)

for i in range(0, test_cnt):
    cur_result = 0
    for i in range(0, num_cnt):
        if(get_random() % (i+1) == 0):
            cur_result = i
    # print(cur_result)
    cnt_list[cur_result] += 1

stand = 1.0/num_cnt
print(stand)

for i in range(0, num_cnt):
    cur_num = float(cnt_list[i])/test_cnt
    print(i, ":\t", cur_num, "\t", cur_num-stand)