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LeetCode.1103.Distribute Candies to People 等差数列分糖果

题目描述

1103 Distribute Candies to People
https://leetcode-cn.com/problems/distribute-candies-to-people/

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


解题过程

暴力法

隐约感觉到是利用等差数列性质,利用公式直接把数组每个位置的元素个数计算出来,但没想清楚具体怎么做。
最后用了暴力法,果然性能很差。
时间复杂度 O(根号c) c为糖果的数量,因为发糖是个等差数列,总和 c = n(n+1)/2,则发糖次数也就是遍历次数n,约等于 根号c
空间复杂度 O(1)

private static class SolutionV2020 {
    public int[] distributeCandies(int candies, int numPeople) {
        if (candies <= 0 || numPeople <= 0) {
            return null;
        }
        int[] ret = new int[numPeople];
        int i = 0;
        while (candies >= 0) {
            ret[i % numPeople] += Math.min(i + 1, candies);
            i++;
            candies -= i;
        }
        return ret;
    }
}

等差数列不等式推导

大体思想是,假设完整发了 p 次糖,由于等差数列 1,…,p 的总和 和 糖果总和 是已知的,根据不等式可以推导出 p 的值,以及最后一次发的不足 p+1 颗糖的数量。
然后只需 O(person_num) 遍历一遍人员数组即可完成分糖。
空间复杂度 O(1)


GitHub代码

algorithms/leetcode/leetcode/_1103_DistributeCandiesToPeople.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_1103_DistributeCandiesToPeople.java


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创建日期 2020-03-05
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