398. Random Pick Index (Medium)

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)

Solution 1: Reservoir Sampling O(n); O(1)

    int[] nums;
    Random rand;

    public RandomPickIndex(int[] nums) {
        this.nums = nums;
        this.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;
            } else if (rand.nextInt(++count) == 0) {
                res = i;
            }
        }
        return res;
    }

Follow Up:

1.find index of maximum number in array, if multiple maximum numbers occurs, return a random index

    public int findIndexOfMax(int[] nums) {
        int res = -1;
        int max = Integer.MIN_VALUE;
        int count = 0;
        Random rand = new Random();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > max) {
                max = nums[i];
                count = 0;
            } else if (nums[i] == max) { //要加else，不然要重复判断
                if (rand.nextInt(++count) == 0) {
                    res = i;
                }
            }
        }
        return res;
    }

    public static void main(String[] args) {
        RandomPickIndex test = new RandomPickIndex();
        int[] nums = {1,2,3,3,3};
        System.out.println(test.findIndexOfMax(nums));
    }