Random Pick Index 398

Question

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

直接的办法就是将所有等于target的值的idnex取出来，再随机取一个。但是这样的空间最坏情况可能为O(n)。

这道题指明了我们不能用太多的空间，那么省空间的随机方法只有水塘抽样了。那么如果了解了水塘抽样，这道题就不算一道难题了，我们定义两个变量，计数器cnt和返回结果res，我们遍历整个数组，如果数组的值不等于target，直接跳过；如果等于target，计数器加1，然后我们在[0,cnt]范围内随机生成一个数字，如果这个数字是0，我们将res赋值为i即可。

代码如下：

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

    public Solution(int[] nums) {
        this.nums = nums;
        this.rnd = new Random();
    }

    public int pick(int target) {
        //this may be O(n) space
        // List<Integer> list = new ArrayList<Integer>();
        // for (int i = 0; i < nums.length; i++) {
        //     if (nums[i] == target) {
        //         list.add(i);
        //     }
        // }
        // if (list.size() == 0) {
        //     return -1;
        // }
        // int index = (int)(Math.random() * list.size());
        // return list.get(index);

        //O(1) space, using Reservoir Sampling
        int res = -1;
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != target) {
                continue;
            }
            if (rnd.nextInt(++count) == 0) {
                res = i;
            }
        }
        return res;
    }
}