398. Random Pick Index

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#LeetCode #Reservoir Sampling #Random Pick Index

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本文介绍了一种在含重复元素的整数数组中随机选取指定目标值索引的方法。利用水库抽样算法，保证每次遇到目标值时，其被选中的概率始终为1/count，确保所有目标值索引具有等概率被返回。

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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);

Reservoir Sampling 的问题。遍历nums，如果nums[i] == target，count++。问题回到了在count个数里面选择一个index，random(count) == 0保证当前数被选到的概率是1/count，再遇到target，count++，依此类推，每次都保证当前数被选到的概率是1/count。代码如下：

public class Solution {
    int[] nums;
    Random random;

    public Solution(int[] nums) {
        this.nums = nums;
        random = new Random();
    }
    
    public int pick(int target) {
        int res = 0;
        int count = 0;
        for (int i = 0; i < nums.length; i ++) {
            if (nums[i] == target) {
                if (random.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);
 */