【Leetcode】398. Random Pick Index

问题分析：

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);

题目的大意是说个给你一个数组，在给你数组中存在的一个数字，你需要随机给出一个该数字在数组中的位置。

分析：

题目说不能用太多的extra space，虽然本来就没想用。想到之前使用的一个蓄水池原理，可以在不知道链表总长度的情况下，等概率的选取链表中的元素。

呃，蓄水池原理各位自行百度吧~

算法概况：时间复杂度O(n)，空间复杂度O(1)

贴上代码：

import java.util.Random;
public class Solution {
    int[] nums;
    int res;
    Random r = new Random();
    public Solution(int[] nums) {
        this.nums = nums;
    }
    
    public int pick(int target) {
        int count=1;
        for(int i=0;i<nums.length;i++){
            if(nums[i] == target){
                if(r.nextInt(count)+1 == count){
                    res = i;
                }
                count++;
            }
        }
        return res;
    }
}


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

贴上结果：

速度还行吧