leetcode: 441. Arranging Coins

最新推荐文章于 2025-10-11 19:00:53 发布

转载最新推荐文章于 2025-10-11 19:00:53 发布 · 155 阅读

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原文链接：https://my.oschina.net/avenger222/blog/837930

本文探讨了LeetCode上编号为441的题目“Arranging Coins”的三种算法解决方法，包括直接计算法、二分法搜索和解不等式法。详细介绍了每种方法的实现代码和原理，其中解不等式法的时间复杂度达到了O(1)，提供了高效的解决方案。

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.

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解法1：直接计算：

# -*- coding:utf-8 -*- 
# leetcode solutions
# 441. Arranging Coins
# coding by Dennis Lu
# 2017/02/11

class Solution(object):
    def arrangeCoins(self, n):
        """
        :type n: int
        :rtype: int
        """

        count = 0
        row_point = 2
        total = 1

        while(total <= n):
        	total += row_point
        	row_point += 1
        	count += 1
        return count

解法2：二分法搜索：

# -*- coding:utf-8 -*- 
# leetcode solutions
# 441. Arranging Coins
# coding by Dennis Lu
# 2017/02/11

class Solution(object):
    def arrangeCoins(self, n):
        """
        :type n: int
        :rtype: int
        """

        first = 1
        last = n
        mid = int((n + 1) / 2)
        
        

        while(1):
        	num0 = (1 + mid) * mid * 0.5
        	num1 = (1 + mid + 1) * (mid + 1) * 0.5

        	if((num0 <= n) and (num1 > n)):
        		return mid
        	if(num0 > n):
        		last = mid - 1
        		mid = int((first + last) / 2)
        		continue
        	if(num1 <= n):
        		first = mid + 1
        		mid = int((first + last) / 2)

解法3：解不等式：

# -*- coding:utf-8 -*- 
# leetcode solutions
# 441. Arranging Coins
# coding by Dennis Lu
# 2017/02/11

import math

class Solution(object):
    def arrangeCoins(self, n):
        """
        :type n: int
        :rtype: int
        """

        return int(math.sqrt(2 * n + 0.25) - 0.5)

3的算法时间复杂度O(1)

转载于:https://my.oschina.net/avenger222/blog/837930