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.java
Note:
The array size can be very large. Solution that uses too much extra space will not pass the judge.算法
Example:markdown
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);
理论参考：Reservoir Sampling 蓄水池抽样算法，经典抽样dom
摘自this
https://discuss.leetcode.com/topic/58301/simple-reservoir-sampling-solution
spa
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;
}
}```
Explanation：.net
public int pick(int target) {
int result = -1;
int count = 0; // to record how many targets in the array
for (int i = 0; i < nums.length; i++) {
if (nums[i] != target)
continue;
/*
For the nth target, ++count is n. Then the probability that rnd.nextInt(++count)==0 is 1/n. Thus, the probability that return nth target is 1/n.
For (n-1)th target, the probability of returning it is (n-1)/n * 1/(n-1)= 1/n.
*/
if (rnd.nextInt(++count) == 0)
result = i;
}
return result;
}