LeetCodeMathdistribute candies to people分糖问题

Posted 2020-12-12

题目：

共有candies颗糖果，n=num_people个人，我们会按照以下方法分发糖果：

我们给第一个人一颗糖果，第二个人两颗糖果，依此类推，直到最后一个人给n颗糖果。

然后，我们返回到行的开头，将n +1颗糖果分配给第一个人，将n + 2颗糖果分配给第二个人，依此类推，直到将2 * n颗糖果分配给最后一个人。

重复此过程（每次我们多给一颗糖果，并在到达终点后移到行的开头），直到糖果用完。 最后一个人将收到我们所有剩余的糖果（不一定比以前的人多一颗）。返回一个数组（长度为num_people个糖果，总和为糖果），该数组代表糖果的最终分布。

Example 1:

Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0,0].
On the third turn, ans[2] += 3, and the array is [1,2,3,0].
On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1].

Example 2:

Input: candies = 10, num_people = 3
Output: [5,2,3]
Explanation: 
On the first turn, ans[0] += 1, and the array is [1,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0].
On the third turn, ans[2] += 3, and the array is [1,2,3].
On the fourth turn, ans[0] += 4, and the final array is [5,2,3].

 

Constraints:

• 1 <= candies <= 10^9
• 1 <= num_people <= 1000

解法:

使用give％num_people确定当前的人的位置，min(candies, give+1)是每次分的糖果数量；
将每次分糖果数量增加1（give+=1,candies-=give），直到用完糖果(while candies>0)。

    def distributeCandies(self, candies: int, num_people: int) -> List[int]:
        people = num_people * [0]
        give = 0
        while candies > 0:
            people[give % num_people] += min(candies, give + 1)
            give += 1
            candies -= give
        return people

Time: O(sqrt(candies)), space: O(num_people).

Runtime: 48 ms, faster than 23.90% of Python3 online submissions for Distribute Candies to People.

Memory Usage: 14.1 MB, less than 19.03% of Python3 online submissions for Distribute Candies to People.