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:

class Solution {
    int [] nums;
    Random rand;
    
    public Solution(int[] nums) {
        this.nums = nums;
        rand = new Random();
    }
    
    public int pick(int target) {
        int res = -1;
        int count = 0;
        for(int i = 0; i<nums.length; i++){
            if(nums[i] != target){
                continue;
            }
            
            if(rand.nextInt(++count) == 0){
                res = i;
            }
        }
        
        return res;
    }
}

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

类似Linked List Random Node, Random Pick with Weight.