0
点赞
收藏
分享

微信扫一扫

【LeetCode】441. Arranging Coins 解题报告(Python)


id: fuxuemingzhu

个人博客: ​​​http://fuxuemingzhu.cn/​​​


目录


  • ​​题目描述​​
  • ​​题目大意​​
  • ​​解题方法​​

  • ​​模拟计算​​
  • ​​二分查找​​
  • ​​数学公式​​

  • ​​日期​​


题目地址:​​https://leetcode.com/problems/arranging-coins/#/description​​

题目描述

You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.

Given n, find the total number of full staircase rows that can be formed.

n is a non-negative integer and fits within the range of a 32-bit signed integer.

Example 1:

n = 5

The coins can form the following rows:

¤
¤ ¤
¤ ¤

Because the 3rd row is incomplete, we return 2.

Example 2:

n = 8

The coins can form the following rows:

¤
¤ ¤
¤ ¤ ¤
¤ ¤

Because the 4th row is incomplete, we return 3.

题目大意

给了n个硬币,要求第k层有k个硬币,问能摆出多少层,如果最后一层不满足的话,是不算的。

解题方法

模拟计算

如果模拟这个安排硬币的过程的话,可以这么做:

class Solution(object):
def arrangeCoins(self, n):
"""
:type n: int
:rtype: int
"""
level = 0
count = 0
while count + level + 1 <= n:
level += 1
count += level
return level

二分查找

上面做法的效率不高。因为前k层的硬币个数可以直接通过(k + 1) * k / 2求出来,所以很直接的想法就可以使用二分查找。

即目的是找到(k + 1) * k / 2<=sum的最大数字,套用二分查找的模板,很容易写出来。

class Solution(object):
def arrangeCoins(self, n):
"""
:type n: int
:rtype: int
"""
left, right = 0, n + 1 #[left, right)
while left < right:
mid = left + (right - left) / 2
if mid * (mid + 1) / 2 <= n:
left = mid + 1
else:
right = mid
return left - 1

数学公式

刷题的时候第一次遇到纯数学的问题,其实就是求解​​sum = (x + 1) * x / 2​​​这个方程,很简单的就能得到​​x = (-1 + sqrt(8 * n + 1)) / 2​​,向下取整就能得到结果了。

下面的代码的括号也要重视一下。

public class Solution {
public int arrangeCoins(int n) {
return (int)((-1 + Math.sqrt(1 + 8 * (long) n)) / 2);
}
}

日期

2017 年 5 月 6 日

2018 年 11 月 24 日 —— 周六快乐



举报

相关推荐

0 条评论