138.Arranging Coins

Posted 2021-01-04 chanaichao

题目：

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.

您想要以楼梯形状形成总共n个硬币，其中每个第k行必须具有恰好k个硬币。

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

给定n，找到可以形成的完整楼梯行的总数。

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

n是一个非负整数，适合32位有符号整数的范围。

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.

解答：

方法一：时间复杂度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：时间复杂度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 }

详解：

方法一：暴力解法

 curr：当前行数也是当前硬币个数

remain：剩余硬币数

从第1行开始遍历，用剩余硬币数减去行数，如果剩余的硬币无法满足下一行需要的硬币数，终止

方法二：二分搜索法

找到前k行之和刚好大于n，k-1即为解

方法三：等差数列性质

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