LeetCode398 随机数索引

题目

Given an integer array nums 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.
Implement the Solution class:
Solution(int[] nums) Initializes the object with the array nums.
int pick(int target) Picks a random index i from nums where nums[i] == target. If there are multiple valid i's, then each index should have an equal probability of returning.

 Example 1: 
Input
["Solution", "pick", "pick", "pick"]
[[[1, 2, 3, 3, 3]], [3], [1], [3]]
Output
[null, 4, 0, 2]

Explanation
Solution solution = new Solution([1, 2, 3, 3, 3]);
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each 
index should have equal probability of returning.
solution.pick(1); // It should return 0. Since in the array only nums[0] is 
equal to 1.
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each 
index should have equal probability of returning.

Constraints:
1 <= nums.length <= 2 * 10⁴
-2³¹ <= nums[i] <= 2³¹ - 1
target is an integer from nums.
At most 10⁴ calls will be made to pick.

方法

哈希表法

• 时间复杂度：O(n)
• 空间复杂度：O(n)

class Solution {
    private Map<Integer,List<Integer>> map = new HashMap<>();
    private Random random = new Random();
    public Solution(int[] nums) {
        for(int i=0;i<nums.length;i++){
            List<Integer> arr = map.getOrDefault(nums[i],new ArrayList<Integer>());
            arr.add(i);
            map.put(nums[i],arr);
        }
    }
    public int pick(int target) {
        List<Integer> arr = map.get(target);
        if(arr!=null){
            int i = random.nextInt(arr.size());
            return arr.get(i);
        }
        return -1;
    }
}

水塘抽样法

• 时间复杂度：O(n)
• 空间复杂度：O(1)

class Solution {
    private int[] nums ;
    private Random random = new Random();
    public Solution(int[] nums) {
        this.nums = nums;
    }
    public int pick(int target) {
        int ans = 0;
        for(int i=0,count=0;i<nums.length;i++){
            if(nums[i]==target){
                count++;
                if(random.nextInt(count)==0){
                    ans = i;
                }
            }
        }
        return ans;
    }
}