Leetcode/F家 -- 398. Random Pick Index(Desgin + Random)

398. Random Pick Index(Design + Random)
Medium
https://leetcode.com/problems/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： use random to ensure equal possibility, key point is save extra space.
解法1: for easier understanding, we create arraylist to store all index of target, and count total,
           Then we use random  from [0,total), to equally pick any element in the array 
Complexity: O(N)time O(m)space - # of target in nums

public class Solution {
    int[] nums;
    Random rand;
    public Solution(int[] nums) {
        this.nums = nums;
        this.rand = new Random();
    }
    public int pick(int target) {
        int total = 0;//total index match target
        int res = -1;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                total++;
                list.add(i);
            }
        }
        res = list.get(rand.nextInt(total)); //int in [0,total)
        return res;
    }
}

解法2: To use no extra space, we pick the index in place.
             (Reservoir Sampling) update by 1/total in the loop
             靠前的index一开始选中的几率高，但是后面loop剩的多被replace的几率也高
             举例有5个元素：
             对于第二个index来说，1/2（pick）*2/3(replace)*3/4(replace)*4/5(replace) = 1/5;
             对于最后一个index来说，就是 1/5(pick) = 1/total.

public class Solution {
    int[] nums;
    Random rand;
    public Solution(int[] nums) {
        this.nums = nums;
        this.rand = new Random();
    }
    public int pick(int target) {
        int total = 0;//total index match target
        int res = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                total++;
                int x = rand.nextInt(total);//[0,total)
                if(x == 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);
 */