1 解題思想

這道題可以理解了為給了n個硬幣，然後需要你按照這個規則：
第i層放i個硬幣

那麼，這n個硬幣，能夠完整的擺好多少層，比如說在第五層時只放了3個，那麼完整的擺了4層，輸出4

這道題直接用等差數列求和公式倒推就可以

直接看公式就可以了
/* 數學推導
假設完成K層，一共N個，由等差數列求和公式有：
(1+k)*k/2 = n
一步步推導：
k+k*k = 2*n
k*k + k + 0.25 = 2*n + 0.25
(k + 0.5) ^ 2 = 2*n +0.25
k + 0.5 = sqrt(2*n + 0.25)
k = sqrt(2*n + 0.25) - 0.5
這裡k是個浮點數，將其取為小於k的最大整數就可以
*/

2 原題

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.

3 AC解

public class Solution {
    public int arrangeCoins(int n) {
        /* 數學推導
        (1+k)*k/2 = n
        k+k*k = 2*n
        k*k + k + 0.25 = 2*n + 0.25
        (k + 0.5) ^ 2 = 2*n +0.25
        k + 0.5 = sqrt(2*n + 0.25)
        k = sqrt(2*n + 0.25) - 0.5
        */
 

        return (int) (Math.sqrt(2*(long)n+0.25) - 0.5);
    }
}