You are given an integer array?gifts?denoting the number of gifts in various piles. Every second, you do the following:

• Choose the pile with the maximum number of gifts.

• If there is more than one pile with the maximum number of gifts, choose any.

• Leave behind the floor of the square root of the number of gifts in the pile. Take the rest of the gifts.

Return?the number of gifts remaining after?k?seconds.

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Example 1:

Input: gifts = [25,64,9,4,100], k = 4Output: 29Explanation: The gifts are taken in the following way: - In the first second, the last pile is chosen and 10 gifts are left behind. - Then the second pile is chosen and 8 gifts are left behind. - After that the first pile is chosen and 5 gifts are left behind. - Finally, the last pile is chosen again and 3 gifts are left behind. The final remaining gifts are [5,8,9,4,3], so the total number of gifts remaining is 29.

Example 2:

Input: gifts = [1,1,1,1], k = 4Output: 4Explanation: In this case, regardless which pile you choose, you have to leave behind 1 gift in each pile. That is, you can't take any pile with you. So, the total gifts remaining are 4.

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Constraints:

• 1 <= gifts.length <= 103

• 1 <= gifts[i] <= 109

• 1 <= k <= 103

Accepted

20,467

Submissions

31,124

1：寫一個函數(shù)返回最大值的index；

2：每次對最大值進行開平方根；

3：最后求數(shù)組的和，返回即可；

用的是普通的遍歷，easy題目了；

Runtime:?8 ms, faster than?27.13%?of?Java?online submissions for?Take Gifts From the Richest Pile.

Memory Usage:?41.4 MB, less than?99.19%?of?Java?online submissions for?Take Gifts From the Richest Pile.