[LeetCode] 398. Random Pick Index ☆☆☆

 

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);

 

解法：

　　因为限制了空间，只能选择省空间的随机方法——水塘采样了。咱们定义两个变量，计数器count和返回结果result，咱们遍历整个数组，若是数组的值不等于target，直接跳过；若是等于target，count加1，而后咱们在[0,count)范围内随机生成一个数字，若是这个数字是0（几率为1/count，知足条件），咱们将result赋值为i便可，参见代码以下：

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

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