leetcode 1103. Distribute Candies to People 解法 python

一.问题描述

We distribute some number of candies, to a row of n = num_people people in the following way:

We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person.

Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person.

This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies.  The last person will receive all of our remaining candies (not necessarily one more than the previous gift).

Return an array (of length num_people and sum candies) that represents the final distribution of candies.

 

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

二.解题思路

可以关注到，其实蜡烛到分配量每次都是递增一个，只是分配的索引再超出num_people的时候要重新开始。

因此我们只需要在索引超出num_people的时候将它求余回到开始，然后蜡烛分配数量每次+1就好了。

还有一点需要注意一下就是那个蜡烛如果不够分就全给了。

减号-操作符的优先级要高于>。

更多leetcode算法题解法请关注我的专栏leetcode算法从零到结束或关注我

欢迎大家一起套路一起刷题一起ac。

三.源码

class Solution:
    def distributeCandies(self, candies: int, num_people: int) -> List[int]:
        rst,i,t=[0]*(num_people+1),0,0
        while candies>0:
            k=i%num_people+1
            num_dst=i+1
            rst[k]=rst[k]+num_dst if candies>=num_dst else rst[k]+candies
            candies-=num_dst  
            i+=1
        return rst[1:num_people+1]