LeetCode 398. Random Pick Index

原题链接在这里：https://leetcode.com/problems/random-pick-index/

题目：

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);

题解：

The question asks no more space, thus try to get pick upon original nums array.

Use Reservoir Sampling: https://www.youtube.com/watch?v=A1iwzSew5QY

Basically, when iterating nums and encounter num == target, accumlate the count and get the random between [0, count).

If random == 0, then update the res index to current index.

Time Complexity: pick, O(n).

Space: O(1). No more array space, since it is pointing to given nums.

AC Java:

 1 class Solution {
 2     int [] nums;
 3     Random rand;
 4     
 5     public Solution(int[] nums) {
 6         this.nums = nums;
 7         rand = new Random();
 8     }
 9     
10     public int pick(int target) {
11         int res = -1;
12         int count = 0;
13         for(int i = 0; i<nums.length; i++){
14             if(nums[i] != target){
15                 continue;
16             }
17             
18             if(rand.nextInt(++count) == 0){
19                 res = i;
20             }
21         }
22         
23         return res;
24     }
25 }
26 
27 /**
28  * Your Solution object will be instantiated and called as such:
29  * Solution obj = new Solution(nums);
30  * int param_1 = obj.pick(target);
31  */

We could also use a HashMap to maintain the num and list of indices.

Get the list of indices based on target and return a random index.

Time Complexity: O(n). n = nums.length.

Space: O(n).

AC Java:

 1 class Solution {
 2     HashMap<Integer, List<Integer>> hm;
 3     Random rand;
 4     public Solution(int[] nums) {
 5         rand = new Random();
 6         hm = new HashMap<>();
 7         for(int i = 0; i < nums.length; i++){
 8             hm.putIfAbsent(nums[i], new ArrayList<Integer>());
 9             hm.get(nums[i]).add(i);
10         }
11     }
12 
13     public int pick(int target) {
14         List<Integer> ids = hm.get(target);
15         return ids.get(rand.nextInt(ids.size()));
16     }
17 }
18 
19 /**
20  * Your Solution object will be instantiated and called as such:
21  * Solution obj = new Solution(nums);
22  * int param_1 = obj.pick(target);
23  */