[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:

  由于限制了空间,只能选择省空间的随机方法——水塘采样了。我们定义两个变量,计数器count和返回结果result,我们遍历整个数组,如果数组的值不等于target,直接跳过;如果等于target,count加1,然后我们在[0,count)范围内随机生成一个数字,如果这个数字是0(概率为1/count,满足条件),我们将result赋值为i即可,参见代码如下:

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);
 */

 

posted @ 2017-02-21 17:29  Strugglion  阅读(266)  评论(0编辑  收藏  举报