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);

思路：一种是Reservoir sampling，也就是在target的count内进行random，如果 == 0，则取这个index；

class Solution {
    private int[] A;
    private Random random;
    public Solution(int[] nums) {
        this.A = nums;
        this.random = new Random();
    }
    
    public int pick(int target) {
        int count = 0;
        int candidate = 0;
        for(int i = 0; i < A.length; i++) {
            if(A[i] == target && random.nextInt(++count) == 0) {
                candidate = i;
            }
        }
        return candidate;
    }
}

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

思路2： 收集target的list，然后返回random index in the list;

class Solution {

    HashMap<Integer, List<Integer>> hashmap;
    public Solution(int[] nums) {
        this.hashmap = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            hashmap.putIfAbsent(nums[i], new ArrayList<>());
            hashmap.get(nums[i]).add(i);
        }
    }
    
    public int pick(int target) {
        List<Integer> indexlist = hashmap.get(target);
        Random random = new Random();
        int index = random.nextInt(indexlist.size());
        return indexlist.get(index);
    }
}

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