Leetcode-398. Random Pick Index

最新推荐文章于 2021-04-15 17:31:20 发布

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本文介绍了一种使用蓄水池算法解决随机选取含重复元素数组中目标值索引的问题，并提供了一个高效的Java实现方案。该算法能在一次遍历内完成任务，适用于大规模数据集。

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。这次比赛略无语，没想到前3题都可以用暴力解。

博客链接：mcf171的博客

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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);

这是一个非常经典的蓄水池算法。 Your runtime beats 91.32% of java submissions.

public class Solution {

    int[] nums;
    Random rnd;

    public Solution(int[] nums) {
        this.nums = nums;
        this.rnd = new Random();
    }
    
    public int pick(int target) {
        int result = -1;
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != target)
                continue;
            if (rnd.nextInt(++count) == 0)
                result = i;
        }
        
        return result;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(nums);
 * int param_1 = obj.pick(target);
 */