398. Random Pick Index

2018-02-20  本文已影响0人  Jeanz

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);

解法:Reservoir Sampling

public class ReservoirSamplingTest {

    private int[] pool; // 所有数据
    private final int N = 100000; // 数据规模
    private Random random = new Random();

    @Before
    public void setUp() throws Exception {
        // 初始化
        pool = new int[N];
        for (int i = 0; i < N; i++) {
            pool[i] = i;
        }
    }

    private int[] sampling(int K) {
        int[] result = new int[K];
        for (int i = 0; i < K; i++) { // 前 K 个元素直接放入数组中
            result[i] = pool[i];
        }

        for (int i = K; i < N; i++) { // K + 1 个元素开始进行概率采样
            int r = random.nextInt(i + 1);
            if (r < K) {
                result[r] = pool[i];
            }
        }

        return result;
    }

    @Test
    public void test() throws Exception {
        for (int i : sampling(100)) {
            System.out.println(i);
        }
    }
}
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