398. Random Pick Index
阿新 • • 發佈:2017-10-25
pan i++ list param pick space add [0 int
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);
含義:給定一個可能包含重復元素的整數數組,給定一個目標數,隨機輸出其下標。你可以假設給定的目標數一定存在於數組中。
註意:數組長度可能很大。使用較多額外空間的解決方案無法通過系統測試。
1 class Solution { 2 3 private int[] nums; 4 public Solution(int[] nums) { 5 this.nums = nums; 6 } 7 8 public int pick(int target) { 9 Random r = new Random();10 ArrayList<Integer> idxs = new ArrayList<Integer>(); 11 for(int i=0;i<nums.length;i++){ 12 if(target==nums[i]){ 13 idxs.add(i); 14 } 15 } 16 return idxs.get(r.nextInt(idxs.size())); 17 } 18 } 1920 /** 21 * Your Solution object will be instantiated and called as such: 22 * Solution obj = new Solution(nums); 23 * int param_1 = obj.pick(target); 24 */
398. Random Pick Index