标签：sum col must += rand continue amp dup possible
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);
class Solution { private int[] arr; private Random rand; public Solution(int[] nums) { this.arr = nums; this.rand = new Random(); } public int pick(int target) { int count = 0; int res = -1; for (int i = 0; i < arr.length; i++) { if (arr[i] != target) { continue; } count += 1; int randNum = rand.nextInt(count); if (randNum == 0) { res = i; } } return res; } } /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(nums); * int param_1 = obj.pick(target); */
标签：sum col must += rand continue amp dup possible
原文地址：https://www.cnblogs.com/xuanlu/p/12181760.html
