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.

您想要以樓梯形狀形成總共n個硬幣，其中每個第k行必須具有恰好k個硬幣。

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

給定n，找到可以形成的完整樓梯行的總數。

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

n是一個非負整數，適合32位有符號整數的範圍。

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.

解答：

方法一：時間復雜度O（n）

 1 class Solution {
 
 2     public int arrangeCoins(int n) {
 3         if(n==0) return n;
 4         int curr=1,remain=n-1;
 5         while(remain>=curr+1){
 6             curr++;
 7             remain-=curr;
 8         }
 9         return curr;
10     }
11 }

方法二：時間復雜度O（logn）

 1 class Solution {
 2     public int arrangeCoins(int
 
 n) {
 3         if(n<=1) return n;
 4         long low=1,high=n;
 5         while(low<high){
 6             long mid=low+(high-low)/2;
 7             if(mid*(mid+1)/2<=n) low=mid+1;
 8             else high=mid;
 9         }
10         return (int)low-1;
11     }
12 }

方法三：

1 class Solution {
2     public int arrangeCoins(int n) {
3         return (int)((-1+Math.sqrt(1+8*(long)n))/2);
4     }
5 }

詳解：

方法一：暴力解法

curr：當前行數也是當前硬幣個數

remain：剩余硬幣數

從第1行開始遍歷，用剩余硬幣數減去行數，如果剩余的硬幣無法滿足下一行需要的硬幣數，終止

方法二：二分搜索法

找到前k行之和剛好大於n，k-1即為解

方法三：等差數列性質

n = (1 + x) * x / 2,得 x = (-1 + sqrt(8 * n + 1)) / 2, 取整

138.Arranging Coins