398. 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);

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); */