題目
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:
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¤ ¤
¤ ¤
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.
題解
這題剛開始用遞推，but超時了。
後來想了想，發現了數學方法：
m階所需要的coins爲m(m+1)/2;現有m個硬幣，我們可以假設x：
x(x+1) /2 = n;
解一元二次方程，對正根取整，得到能夠得到的stair number;
這裏很多人不理解爲啥向下取整就可以了。其實我們可以發現，假設正根x1向下取整爲k,則滿足：
k小於等於x1小於k+1;（不能打小於號，也是醉了）
那麼當coins爲k*（k+1）/2時的stair 數目是k,當coins爲(k+1）(k+2)/2時，stair 數目是k+1;
因此對於所有介於k*（k+1）/2與(k+1）(k+2)/2的所有n值，stair number =k;
那麼此時的k值就是能夠達到的stair number；
class Solution {
public:
int arrangeCoins(int n) {
return sqrt(2 * (long)n + 1 / 4.0) - 1 / 2.0;
}
};做完這題發現可以回初中重新讀書了（表情都發不出來）；